Solutions Manual_Chapter 11_Final.pdf
Transcript of Solutions Manual_Chapter 11_Final.pdf
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 1/135
a
Ans.
Ans.T2 = 00
- + T2 = 0+ T ©Fy = 0;
T1 =
- T1 (75) = 0+ ©MC = 0; 225(375)
1125 N
225 1125
9 N
-
11–1. The load binder is used to support a load. If the force
applied to the handle is 225 N, determine the tensions T 1 and
T 2 in each end of the chain and then draw the shear and
moment diagrams for the arm ABC .
300 mm
225 N
T2
T1
75 mm
A
B
C
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1047
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 2/135
11–2. Draw the shear and moment diagrams for the shaft.The bearings at A and D exert only vertical reaction on theshaft.The loading is applied to the pulleys at B and C and E.
A
B
350 mm 500 mm 375 mm 300 mm
360 N
500 N
160 N
C D
E
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1048
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 3/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
a
0;
3
5 (4000) 0;
2400 lb
0;
4
5 (4000) 1200 0;
2000 lb
4000 lb 0; 4
5 (3) 1200(8) 0;
The engine crane is used to support the engine, whichhas a weight of 1200 lb. Draw the shear and moment diagramsof the boom ABC when it is in the horizontal position shown.
5 ft3 ft
C B
4 ft
Ahas a weight of 6 kN. Draw the shear and moment diagrams
F B(0.9) – 6(2.4) = 0 F B = 20 kN
(20) – 6 = 0; A y = 10 kN
(20) = 0; A x = 12 kN
1.5 m0.9 m
C B
1.2 m
A
0.9 m 1.5 m
F B 6 kN
V kN
6
–10
–9
14 kN-m
11–3.
1049
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 4/135
*11–4. Draw the shear and moment diagrams for the beam.
1.2 m 1.2 m 1.2 m 1.2 m 1.2 m
9 kN 9 kN 9 kN 9 kN
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1050
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 5/135
11–5. Draw the shear and moment diagrams for the beam.
2 m 3 m
10 kN 8 kN
15 kNm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1051
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 6/135
11–6. Express the internal shear and moment in terms of x
and then draw the shear and moment diagrams.
A B
x
L
2
L
2
w0
Support Reactions: Referring to the free-body diagram of the entire beamshown in Fig. a,
a
Shear and Moment Function: For we refer to the free-body
diagram of the beam segment shown in Fig. b.
Ans.
a
Ans.
For we refer to the free-body diagram of the beam segment
shown in Fig. c.
Ans.
a
Ans.
When V = 0, the shear function gives
Substituting this result into the moment equation,
Shear and Moment Diagrams:As shown in Figs. d and e.
M|x= 0.7041L = 0.0265w0L2
x = 0.7041L0 = L2 - 6(2x - L)2
M = w0
24LcL2x - (2x - L)3 d
M +1
2cw0
L(2x - L) d c1
2(2x - L) d c1
6(2x - L) d -
w0
24Lx = 0+ ©M = 0;
V = w0
24LcL2 - 6(2x - L)2 d
w0L
24 - 1
2 cw0
L(2x - L) d c12 (2x - L) d - V = 0+ c ©Fy = 0;
L
2 6 x … L,
M = w0L
24 x
M - w0L
24 x = 0+ ©M = 0;
V = w0L
24
w0L
24 - V = 0+ c ©Fy = 0;
0 … x 6 L
2,
Ay = w0L
24
Ay +5
24 w0L -
1
2 w0aL2 b = 0+ c ©Fy = 0;
By =5
24 w0L
By(L) -1
2 w0
aL
2 b a5
6L
b = 0+ ©MA = 0;
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1052
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 7/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
4 ft
6 kip 8 kip
A
C
B
6 ft 4 ft 4 ft30 kN
40 kN
20 kN20 kN
20 kN50 kN
1.5 m1 m
1 m 1 m
1.5 m
1 m
20
20
–30
–30
–20
1 m
30 kN 40 kN
A
C
B
1.5 m 1 m 1 m
11–7. Draw the shear and moment diagrams for thecompound beam which is pin connected at B. (This structureis not fully stable. But with the given loading, it is balancedand will remain as shown if not disturbed.)
1053
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 8/135
Support Reactions: Referring to the free-body diagram of the entire beam shownin Fig. a,
a
Shear and Moment Function: For , we refer to the free -body diagramof the beam segment shown in Fig. b.
lb Ans.
a
Ans.
For we refer to the free-body diagram of the beam segment shownin Fig. c.
Ans.
a
Ans.
Shear and Moment Diagrams:As shown in Figs. d and e.
M = { ( - x)} N#( - x) - M = 0+ ©M = 0;
V = -
V + = 0+ c ©Fy = 0;
6 x … 3 m,
M = { x - 00x2} #
M + xax2b - x = 0+ ©M = 0;
V = {9333 - x}
- x - V = 0+ c ©Fy = 0;
0 … x 6
Ay =
Ay + - 00( ) - = 0+ c ©Fy = 0;
By =
By(3) - 00 ( )( ) - 00( ) = 0+ ©MA = 0;
*11–8. Express the internal shear and moment in terms of xand then draw the shear and moment diagrams for the beam.
40 26 0 2 1
6 67 N6
60 26667 4000
9333 N
2 m
6000
60009333
93336000
N m039333
2 m
6667
6667 N
6667 3
m6667 3
A
B
4000 N
6 kN/m
2 m 1 m
x
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1054
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 9/135
11–9. Express the internal shear and moment in terms of x
and then draw the shear and moment diagrams for theoverhanging beam.
Support Reactions: Referring to the free-body diagram of the entire beam shownin Fig. a,
a
Shear and Moment Function: For we refer to the free-body diagramof the beam segment shown in Fig.b.
Ans.
a
Ans.
For we refer to the free-body diagram of the beam segment shownin Fig. c.
Ans.
a
Ans.
Shear and Moment Diagrams:As shown in Figs. d and e.
M = -{3(6 - x)2} kN # m
-M - 6(6 - x)a6 - x2 b = 0+ ©M = 0;
V = {6(6 - x)} kN
V - 6(6 - x) = 0+ c ©Fy = 0;
4 m 6 x … 6 m,
M = {9x - 3x2} kN # m
M + 6xax2b - 9x = 0+ ©M = 0;
V = {9 - 6x} kN
9 - 6x - V = 0+ c ©Fy = 0;
0 … x 6 4 m,
Ay = 9 kN
Ay + 27 - 6(6) = 0+ c ©Fy = 0;
By = 27 kN
By(4) - 6(6)(3) = 0+ ©MA = 0;
A
B
6 kN/ m
x
4 m 2 m
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1055
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 10/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
Equations of Equilibrium: Referring to the free-body diagram of the frame shownin Fig. a,
a
Shear and Moment Diagram: The couple moment acting on B due to N D
is
. The loading acting on member ABC is shown in Fig. band the shear and moment diagrams are shown in Figs.c and d. 300(1.5) 450 lb ft
300 lb
0;
(1.5) 150(3) 0
150 lb
0;
150 0
11–10. Members ABC and BD of the counter chair arerigidly connected at B and the smooth collar at D is allowedto move freely along the vertical slot. Draw the shear andmoment diagrams for member ABC.
A
D
B
C
P 150 lb
1.5 ft1.5 ft
1.5 ft
A
D
B
C
P 750 N
0.45 m0.45 m
0.45 m
750
750 N
N D = (0.45) – 750(0.9) = 0
N D = 1500 N
M B = 1500(0.45) = 675 N · m. The loading acting on member ABC is shown in Fig. b0.45 m 0.45 m
750 N
0.45 m
0.45 m 0.45 m750 N
M B = 675 N· m1500 N
V (N)
750
0.45 0.9
x (m)0.45
–337.5
0.9
337.5
M (N · m)
1056
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 11/135
11–11. Draw the shear and moment diagrams for the pipe.The end screw is subjected to a horizontal force of 5 kN.Hint: The reactions at the pin C must be replaced by anequivalent loading at point B on the axis of the pipe. 80 mm
400 mm
5 kN
A
B
C
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1057
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 12/135
*11–12. A reinforced concrete pier is used to support thestringers for a bridge deck. Draw the shear and momentdiagrams for the pier when it is subjected to the stringerloads shown. Assume the columns at A and B exert onlyvertical reactions on the pier.
1 m 1 m 1 m 1 m1.5 m
60 kN 60 kN35 kN 35 kN 35 kN1.5 m
A B
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1058
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 13/135
11–13. Draw the shear and moment diagrams for the rod. Itis supported by a pin at A and a smooth plate at B.The plateslides within the groove and so it cannot support a verticalforce,although it can support a moment.
4 m
A
B
2 m
15 kN
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1059
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 14/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
10 in.4 in.
50 in. A B
C
D
120
6–14. The industrial robot is held in the stationary positionshown.Draw the shear and moment diagrams of the arm ABC
if it is pin connected at A and connected to a hydraulic cylinder(two-force member) BD. Assume the arm and grip have auniform weight of 1.5 lbin. and support the load of 40 lb atC .
250 mm100 mm
1250 mm A B
C
D
120
11–14. The industrial robot is held in the stationary positionshown. Draw the shear and moment diagrams of the arm
ABC if it is pin connected at A and connected to a hydrauliccylinder (two-force member) BD. Assume the arm and griphave a uniform weight of 0.3 N/mm and support the load of200 N at C .
200 N250 mm
100 mm1469 N
0.3 N/mm
1250 mm
1469 N
1864 N
2544 NV (N)
575
200
–30
–1969–1894
–484.4
–1.5
M (N · m)
1060
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 15/135
*11–16. Determine the placement distance a of the rollersupport so that the largest absolute value of the moment isa minimum. Draw the shear and moment diagrams for thiscondition.
Support Reactions: As shown on FBD.
Absolute Minimum Moment: In order to get the absolute minimum moment, themaximum positive and maximum negative moment must be equal that is
For the positive moment:
a
For the negative moment:
a
Ans.
Shear and Moment Diagram:
a = 2 3
2 L = 0.866L
4a L - 3L2 = 4a L - 4a
2
PL -3PL
2
4a= P(L - a)
Mmax( + ) = Mmax( - )
Mmax( - ) = P(L - a)
Mmax( - ) - P(L - a) = 0+ ©MNA = 0;
M max( + ) = PL -3PL
2
4a
Mmax( + ) - a2P -3PL
2a b aL
2 b = 0+ ©MNA = 0;
Mmax( + ) = Mmax( - ).
A
P
a
P
B
L–2
L–2
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1062
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 16/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
(1)
a (2) 0;
1
2
(33.33)()
3 300 0 {300 5.5563} lb ft
0; 300 1
2 (33.33)() 0 {300 16.672} lb
300 lb 200 lb/ ft
A
6 ft
The free-body diagram of the beam’s left segment sectioned through an arbitrarypoint shown in Fig. b will be used to write the shear and moment equations. Theintensity of the triangular distributed load at the point of sectioning is
Referring to Fig.b,
w 2006 33.33
The shear and moment diagrams shown in Figs. c and d are plotted using Eqs. (1)and (2), respectively.
V (kN)
M (kN · m)
–1.5
–4.5
x (m)
x (m)
–5
1.5 kN
4.5 kN
5 kN · m
12
(3)(2) kN
43
m 23
m
12
(1.5 x) x
1.5 kN 3 kN/m
A
2 m
3 x2 = 1.5 x
–1.5 –1
2 (1.5 x)( x) – V = 0 V = {–1.5 – 0.75 x2} kN (1)
(1.5 x)( x) x3
+ 1.5 = 0 M = {–1.5 – 0.25 x3} kN · m (2)
1.5 kN
11–17. Express the internal shear and moment in thecantilevered beam as a function of x and then draw theshear and moment diagrams.
1063
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 17/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
Support Reactions: As shown on FBD.
Shear and Moment Function:
For :
Ans.
a
Ans.
For :
Ans.
a
Ans. {8.00 120} kip ft
0;
8(10 ) 40 0
0; 8 0 8.00 kip
6 ft 10 ft
{2 30.0 216} kip ft
0; 216 22 30.0 0
{30.0 2} kip
0;
30.0 2 0
0 6 ft
11–18. Draw the shear and moment diagrams for the beam,and determine the shear and moment throughout the beamas functions of x.
6 ft 4 ft
2 kip/ ft 8 kip
x
10 kip
40 kipft
1.8 m 1.2 m
30 kN/m40 kN
x
50 kN
200 kNm
458.6 kN · m
30(1.8) = 54 kN50 kN 40 kN
144 kN
200 kN · m
0.9 m 0.9 m 1.2 m
30 x
144 kN
458.6 kN · m
40 kN
200 kN · m
3 – x
V (kN)
14490
1.8 3
40 1.8 3 x (m) x (m)
–248–200
–458.6
M (kN · m)
1.8 m
144 – 30 x – V = 0
V = {144 – 30 x} kN
M + 458.6 + 30 x x2 – 144 x = 0
M = {–15 x2 + 144 x – 458.6} kN · m
For 1.8 m < x 3 m:
V – 40 = 0 V = 40 kN
–M – 40(3 – x) – 200 = 0
M = {40 x – 320} kN · m
1064
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 18/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
A
30 kipft
B
5 ft 5 ft
2 kip/ ft
5 ft
11–19. Draw the shear and moment diagrams for the beam.
30 kN/m 45 kN · m
3.75 kN
–3.75
–45
V (kN)
M (kN · m)
1.5 m 1.5 m 1.5 m
5.625
–39.375–33.75
A
45 kNm
B
1.5 m 1.5 m
30 kN/m
1.5 m41.25 kN
1065
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 19/135
*11–20. Draw the shear and moment diagrams for theoverhanging beam.
Support Reactions: Referring to the free-body diagram of the beam shown in Fig. a,
a
Shear and Moment Functions: For we refer to the free-body diagramof the beam segment shown in Fig. b.
Ans.
a
Ans.
When V = 0, from the shear function,
Substituting this result into the moment function,
For we refer to the free-body diagram of the beam segmentshown in Fig. c.
Ans.
a
Ans.
Shear and Moment Diagrams: As shown in Figs. d and e.
M = {40.05(6 - x)}
( - x) - M = 0+ ©M = 0;
V = - k
V + = 0+ c ©Fy = 0;
6 x … 6 m,
M|x= .94 = 9. kN #
x = .940 = - x2
M = e x - x3 f kN #
M + ax3b - x = 0+ ©M = 0;
V =
e - x2
f kN
-1
2a b(x) - V = 0+ c ©Fy = 0;
0 … x 6 ,
Ay = k
Ay + -1
2 ( )( ) = 0+ c ©Fy = 0;
By = k
By(6) -1
2 ( )( )( ) = 0+ ©MA = 0; 45 4 2.67
40.05 N
40.05 45 4
49.95 N
4 m
49.95 1x
49.95
49.95 m
49.95 2 9 m
2 9 m 9 21 m
4 m
40.05kN
40.05 N
40.05 6
11.25
49.95 5.625
1
2a b(x)1x11.25
1.875
5.625
kN # m
B A
4 m
45 kN/m
2 m
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1066
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 20/135
Ans.Mmax = #393.75 N m
11–21. The 700-N man sits in the center of the boat, which
has a uniform width and a weight per linear foot of 45 N/m.
Determine the maximum bending moment exerted on the
boat. Assume that the water exerts a uniform distributed
load upward on the bottom of the boat.
2.25 m 2.25 m
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1067
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 21/135
11–22. Draw the shear and moment diagrams for theoverhang beam.
Since the loading is discontinuous at support B, the shear and moment equations mustbe written for regions and of the beam.The free-bodydiagram of the beam’s segment sectioned through an arbitrary point within these tworegions is shown in Figs. b and c.
Region , Fig. b0 … x 6 3 m
3 m 6 x … 6 m0 … x 6 3 m
4 kN/ m
3 m 3 m
A
B
(1)
a (2)
Region , Fig.c
(3)
a (4)+ ©M = 0; -M - 4(6 - x) c 1
2 (6 - x) d = 0
M = {-2(6 - x)2}kN # m
+ c ©Fy = 0; V - 4(6 - x) = 0 V = {24 - 4x} kN
3 m 6 x … 6 m
+ ©M = 0; M +1
2a4
3xb(x)ax
3b + 4x = 0 M = e -
2
9 x3 - 4xf kN # m
+ c ©Fy = 0; -4 -1
2a 4
3 xb(x) - V = 0 V = e -
2
3 x2 - 4 f kN
The shear diagram shown in Fig. d is plotted using Eqs. (1) and (3). The value of shear just to the left and just to the right of the support is evaluated using Eqs. (1)and (3), respectively.
The moment diagram shown in Fig. e is plotted using Eqs. (2) and (4).The value of
the moment at support B is evaluated using either Eq. (2) or Eq. (4).
or
Mx= 3 m = -2(6 - 3)2 = -18 kN # m
Mx= 3 m = -2
9(33) - 4(3) = -18 kN # m
Vx=3 m + = 24 - 4(3) = 12 kN
Vx= 3 m - = -2
3 (32) - 4 = -10 kN
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1068
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 22/135
11–23. The footing supports the load transmitted by the twocolumns. Draw the shear and moment diagrams for thefooting if the reaction of soil pressure on the footing isassumed to be uniform.
2 m 4 m 2 m
60 kN60 kN
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1069
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 23/135
*11–24. Express the shear and moment in terms of x andthen draw the shear and moment diagrams for the simplysupported beam.
A B
3 m 1.5 m
300 N/ m
Support Reactions: Referring to the free-body diagram of the entire beam shownin Fig. a,
a
Shear and Moment Function: For we refer to the free-body diagram
of the beam segment shown in Fig. b.
Ans.
a
Ans.
When V = 0, from the shear function,
Substituting this result into the moment equation,
For we refer to the free-body diagram of the beam segmentshown in Fig. c.
Ans.
a
Ans.
Shear and Moment Diagrams: As shown in Figs. d and e.
M = e 375(4.5 - x) -100
3(4.5 - x)3f N # m
375(4.5 - x) -1
2 [200(4.5 - x)](4.5 - x)a4.5 - x
3 b - M = 0+ ©M = 0;
V =
e100(4.5 - x)2 - 375
f N
V + 375 -1
23200(4.5 - x)4(4.5 - x) = 0+ c ©Fy = 0;
3 m 6 x … 4.5 m,
M| x = 2 6 m = 489.90 N # m
x = 2 6 m0 = 300 - 50x2
M = e300x -50
3 x3f
N # m
M +1
2 (100x)xax
3b - 300x = 0+ ©M = 0;
V = {300 - 50x2} N
300 -1
2(100x)x - V = 0+ c ©Fy = 0;
0 … x 6 3 m,
Ay = 300 N
Ay + 375 -1
2 (300)(3) -
1
2 (300)(1.5) = 0+ c ©Fy = 0;
By = 375 N
By(4.5) -1
2 (300)(3)(2) -
1
2 (300)(1.5)(3.5) = 0+ ©MA = 0;
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1070
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 24/135
Ans.
a
Ans.M = - 5x2 + x -
- 0 - 50(x - (x -
2 - M + 50(x - ) = 0+ ©M = 0;
V = - 50x
+ c ©Fy = 0;
- 50(x - ) - V + 50 = 01.222 22
4950 22
22 1. 21.2)
22 1.2)
11.2 4950 4950
27
11–25. Draw the shear and moment diagrams for the beam
and determine the shear and moment in the beam as
functions of x, where 1.2 m < x < 3 m.
270 N.m
B
x
1.2 m 1.2 m
2.25 kN/m
2 m
270 N.m
A
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1071
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 25/135
11–27. Draw the shear and moment diagrams for the beam.
a
Substitute ,
M = 0.0345 w0L2
x = 0.7071L
+ ©MNA = 0; M +1
2 aw0x
L b (x)ax
3b -
w0L
4 ax -
L
3 b = 0
x = 0.7071 L
+ c ©Fy = 0;
w0L
4 -
1
2 aw0x
L b (x) = 0
B
w0
A2L
3
L
3
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1072
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 26/135
Support Reactions: As shown on FBD.
Shear and Moment Diagram: Shear and moment at can be determinedusing the method of sections.
a
M =5w0 L2
54
+ ©MNA = 0; M + w0 L
6 aL
9 b -
w0 L
3 aL
3 b = 0
+ c ©Fy = 0; w0 L
3 -
w0 L
6 - V = 0 V =
w0 L
6
x = L>3
*11–28. Draw the shear and moment diagrams for the beam. w0
A B
L
3
L
3
L
3
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1073
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 27/135
Equations of Equilibrium: Referring to the free-body diagram shown in Fig. a,
a
Shear and Moment Diagram: As shown in Figs. b and c.
Ay = 24 kN
Ay + 24 - 8(6) = 0+ c ©Fy = 0;
By = 24 kN
By(3) - 8(6)(1.5) = 0+ ©MA = 0;
11–29. Draw the shear and moment diagrams for the doubleoverhanging beam.
B A
1.5 m 1.5 m3 m
8 kN/ m
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1074
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 28/135
11–30. The beam is bolted or pinned at A and rests on a bearing
pad at B that exerts a uniform distributed loading on the beam
over its 0.6-m length. Draw the shear and moment diagrams
for the beam if it supports a uniform loading of 30 kN/m.
2.4 m
0.3 m 0.6 m
A
B
30 kN/m
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1075
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 29/135
11–31. The support at A allows the beam to slide freely alongthe vertical guide so that it cannot support a vertical force.Draw the shear and moment diagrams for the beam.
B A
L
w
Equations of Equilibrium: Referring to the free-body diagram of the beam shownin Fig. a,
a
Shear and Moment Diagram: As shown in Figs. b and c.
By = wL
By - wL = 0+ c ©Fy = 0;
MA = wL2
2
wLaL
2 b - MA = 0+ ©MB = 0;
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1076
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 30/135
Ans.w0 = 1.2 kN>m
+ c ©Fy = 0;
2(w0)(20)a 1
2b - 60(0.4) = 0
*11–32. The smooth pin is supported by two leaves A and Band subjected to a compressive load of caused bybar C . Determine the intensity of the distributed load of the leaves on the pin and draw the shear and momentdiagram for the pin.
w0
0.4 kN>m
20 mm
0.4 kN/ m
w0
20 mm 60 mm
w0 A B
C
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1077
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 31/135
Equations of Equilibrium: Referring to the free-body diagram of the shaft shownin Fig. a,
a
Shear and Moment Diagram: As shown in Figs. b and c.
Ay = 650 N
Ay + 250 - 900 = 0+ c ©Fy = 0;
By = 250 N
By(2) + 400 - 900(1) = 0+ ©MA = 0;
11–33. The shaft is supported by a smooth thrust bearing at A and smooth journal bearing at B. Draw the shear andmoment diagrams for the shaft.
A B
1 m 1 m 1 m
900 N
400 Nm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1078
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 32/135
Equations of Equilibrium: Referring to the free-body diagram of the beam shown inFig. a,
a
Shear and Moment Diagram: As shown in Figs. b and c.
MA = 9 kN # m
MA - 3 - 2(3) = 0+ ©MA = 0;
Ay = 2 kN
Ay - 2 = 0+ c ©Fy = 0;
11–34. Draw the shear and moment diagrams for thecantilever beam.
A
2 kN
3 kNm
1.5 m 1.5 m
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1079
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 33/135
Support Reactions: As shown on FBD.
Shear and Moment Functions:
For :
Ans.
a
Ans.
For :
Ans.
Set ,
a
Ans.
Substitute , M = 691 N # mx = 3.87 m
M =
e -
100
9
x3 + 500x - 600
f N # m
+ 200(x - 3)ax - 3
2 b - 200x = 0
+ ©MNA = 0; M +1
2 c 200
3 (x - 3) d(x - 3)ax - 3
3 b
x = 3.873 mV = 0
V = e -
100
3 x2 + 500f N
+ c ©Fy = 0;
200 - 200(x - 3) -1
2 c200
3 (x - 3) d(x - 3) - V = 0
3 m 6 x … 6 m
M = 5200 x6 N # m
+ ©MNA = 0; M - 200 x = 0
+ c ©Fy = 0; 200 - V = 0 V = 200 N
0 … x 6 3 m
11–35. Draw the shear and moment diagrams for the beamand determine the shear and moment as functions of x.
3 m 3 m
x
A B
200 N/ m
400 N/ m
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1080
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 34/135
*11–36. The shaft is supported by a smooth thrust bearing at A and a smooth journal bearing at B. Draw the shear andmoment diagrams for the shaft.
Equations of Equilibrium: Referring to the free-body diagram of the shaft shown inFig. a,
a
Shear and Moment Diagram:As shown in Figs. b and c.
Ay = 825 N
1725 - 900 - Ay = 0+ c ©Fy = 0;
By = 1725 N
By(1.6) - 600 - 900(2.4) = 0+ ©MA = 0;
A B
900 N
600 Nm
0.8 m 0.8 m 0.8 m
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1081
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 35/135
11–37. Draw the shear and moment diagrams for the beam.
B
4.5 m 4.5 m
50 kN/ m
A
50 kN/ m
A
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1082
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 36/135
11–38. The beam is used to support a uniform load along CDdue to the 6-kN weight of the crate. If the reaction atbearing support B can be assumed uniformly distributedalong its width, draw the shear and moment diagrams forthe beam.
2.75 m 2 m
0.75 m0.5 m
C
B
A
D
Equations of Equilibrium: Referring to the free-body diagram of the beamshown in Fig. a,
a
Shear and Moment Diagram: The intensity of the distributed load at
support B and portion CD of the beam are
Fig. b. The shear
and moment diagrams are shown in Figs.c and d.
= 3 kN>m,20 kN>m and wCD =6
2
100.5
=wB = FB0.5
=
Ay = 4 kN
10 - 6 - Ay = 0+ c ©Fy = 0;
FB = 10 kN
FB(3) - 6(5) = 0+ ©MA = 0;
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1083
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 37/135
Equations of Equilibrium: Referring to the free-body diagram of the beam shown inFig. a,
a
Shear and Moment Diagram: As shown in Figs. b and c.
Ay =
Ay + 00 - 0 - 00( ) - 00 = 0+ c ©Fy = 0;
By = 00
By( ) + 00( ) - 00( )( ) - 00( ) = 0+ ©MA = 0;
11–39. Draw the shear and moment diagrams for the doubleoverhanging beam.
2 18 1 30 2 1 18 3
48 N
1830 248 180
4800 N
1 m 1 m
3000 N/m
1800 N
2 m
1800 N
B A
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1084
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 38/135
*11–40. Draw the shear and moment diagrams for thesimply supported beam.
AB
2 m 2 m
10 kN 10 kN
15 kNm
2 m
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1085
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 39/135
Support Reactions: Referring to the free-body diagram of segment BC shownin Fig. a,
a
Using the result of B y and referring to the free-body diagram of segment AB, Fig. b,
a
Shear and Moment Diagrams: As shown in Figs. c and d.
MA = 2800 N
MA - 600(2) - 400(4) = 0+ ©MA = 0;
Ay = 1000 N
Ay - 600 - 400 = 0+ c ©F y = 0;
By = 400 N
By + 400 - 400(2) = 0+ c ©F y = 0;
Cy = 400 N
Cy(2) - 400(2)(1) = 0+ ©MB = 0;
11–41. The compound beam is fixed at A, pin connected atB, and supported by a roller at C . Draw the shear andmoment diagrams for the beam.
A BC
2 m2 m2 m
400 N/ m
600 N
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1086
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 40/135
Support Reactions:
From the FBD of segment AB
a
From the FBD of segment BD
a
From the FBD of segment AB
Shear and Moment Diagram:
©Fx = 0; Ax = 0:+
©Fx = 0;
Bx = 0:+
Cy = 20.0 kN
+ c ©Fy = 0; Cy - 5.00 - 5.00 - 10.0 = 0
Dy = 5.00 kN
+ ©MC = 0; 5.00(1) + 10.0(0) - Dy (1) = 0
+ c ©Fy = 0; Ay - 10.0 + 5.00 = 0 Ay = 5.00 kN
+ ©MA = 0;
By (2) - 10.0(1) = 0
By = 5.00 kN
11–42. Draw the shear and moment diagrams for thecompound beam.
B A C D
2 m 1 m 1 m
5 kN/ m
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1087
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 41/135
11–43. The compound beam is fixed at A, pin connected atB, and supported by a roller at C . Draw the shear andmoment diagrams for the beam.
Support Reactions: Referring to the free-body diagram of segment BC shownin Fig. a,
a
Using the result of B y and referring to the free-body diagram of segment AB, Fig.b,
a
Shear and Moment Diagrams: As shown in Figs. c and d.
MA = 19.5 kN # m
MA - 2(3) - 4.5(3) = 0+ ©MA = 0;
Ay = 6.5 kN
Ay - 2 - 4.5 = 0c ©Fy = 0;
By = 4.5 kN
By + 4.5 - 3(3) = 0c ©Fy = 0;
Cy = 4.5 kN
Cy(3) - 3(3)(1.5) = 0+ ©MB = 0;
A BC
3 kN/ m
2 kN
3 m 3 m
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1088
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 42/135
x =
18L
2.4
0
x3dx
106.65 = 1.8 m
FR =1
8L2.4
0
x dx2
= 106.65 kN
*11–44. Draw the shear and moment diagrams for the beam.
2.4 m
A B
x
w
w=1 x
2
120 kN/ m
1–8
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
106.65 kN
kN # m192
kN
106.65
kN #m
- 192
1089
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 43/135
Support Reactions: Referring to the free-body diagram of segment BC shownin Fig. a,
a
Using the result of FB and referring to the free-body diagram of segment AB, Fig.b,
a
Shear and Moment Diagrams: As shown in Figs. c and d.
MA = 54 kN # m
MA -1
2(3)(4.5)(3) - 7.5(4.5) = 0+ ©MA = 0;
Ay = 14.25 kN
Ay -1
2(3)(4.5) - 7.5 = 0+ c ©F y = 0;
Cy = 7.5 kN
Cy + 7.5 - 15 = 0+ c ©F y = 0;
FB = 7.5 kN
15(1.5) - FB(3) = 0+ ©MC = 0;
11–45. A short link at B is used to connect beams AB andBC to form the compound beam shown. Draw the shear andmoment diagrams for the beam if the supports at A and Bare considered fixed and pinned, respectively.
A C B
4.5 m 1.5 m 1.5 m
15 kN
3 kN/ m
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1090
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 44/135
Support Reactions: The intensity of the uniform distributed load caused by its ownweight is Due to symmetry,
Absolute Minimum Moment: To obtain the absolute minimum moment, themaximum positive moment must be equal to the maximum negative moment. The
maximum negative moment occurs at the supports. Referring to the free-bodydiagram of the beam segment shown in Fig. b,
a
The maximum positive moment occurs between the supports. Referring to thefree-body diagram of the beam segment shown in Fig. c,
Using this result,
a
It is required that
Solving for the positive result,
Ans.
Shear and Moment Diagrams: Using the result for b, Fig. d, the shear and momentdiagrams are shown in Figs. e and f .
b = 0.2071L = 0.207L
4b2 + 4Lb - L2 = 0
ga2L
8 (L - 4b) =
ga2b2
2
Mmax( + ) = Mmax( - )
Mmax( + ) = ga2L
8(L - 4b)
Mmax( + ) + ga2aL
2 b aL
4 b -
ga2L
2 aL
2 - bb = 0+ ©M = 0;
x = L
2
ga2L
2 - ga2 x = 0+ c ©F y = 0;
Mmax( - ) = ga2b2
2
ga2bab
2b - Mmax( - ) = 0+ ©M = 0;
Fy = ga2L
2
2Fy - ga2L = 0+ c ©F y = 0;
w = ga2.
11–46. Determine the placement b of the hooks to minimizethe largest moment when the concrete member is beinghoisted. Draw the shear and moment diagrams. Themember has a square cross section of dimension a on eachside.The specific weight of concrete is .g
L
b b
60 60
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1091
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 45/135
11–46. Continued
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1092
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 46/135
Bending Stress-Curvature Relation:
Ans.
Moment Curvature Relation:
Ans.M = 61.875 N # m = 61.9 N # m
1
0.9091 =
M
200(109) c 1
12 (1)(0.00153) d
1
r =
M
EI;
r = 0.9091 m = 909 mm
165(106) =200(109)30.75(10- 3)4
rsallow =
Ec
r ;
11–47. If the A-36 steel sheet roll is supported as shown andthe allowable bending stress is 165 MPa, determine thesmallest radius r of the spool if the steel sheet has a width of 1 m and a thickness of 1.5 mm.Also, find the correspondingmaximum internal moment developed in the sheet.
r
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1093
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 47/135
Applying the flexure formula
Ans.M = 13.53 # m
70 = M (262 - 84.7)
smax = Mc
I
*11–48. Determine the moment M that will produce amaximum stress of 70 MPa on the cross section.
75 mm
D
AB
12 mm
M
12 mm
75 mm
C
250 mm
12 mm12 mm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
Section Properties:
91.73 in4
1
12 (0.5) 103 0.5(10)(5.5 3.40)2
2 1
12 (0.5)(33) 0.5(3)(3.40 2)2
1
12 (4) 0.53 4(0.5)(3.40 0.25)2
0.25(4)(0.5) 2[2(3)(0.5)] 5.5(10)(0.5)
4(0.5) 2[(3)(0.5)] 10(0.5) 3.40 in.
(6)(99)(12) + 2[49.5(75)(12)] + 137(250)(12)
99(12) + 2[(75)(12)] + 250(12) 84.7 mm
I NA 1
12 (99)(123) + 99(12)(84.7 – 6)2
+ 2 1
12 (12)(753) + (12)(75)(84.7 – 49.5)2
+1
12 (12)(2503) + (12)(250)(137 – 84.7)2
34.277(106) mm4
Maximum Bending Stress:
34.277 106×
kN
1094
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 48/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
Section Properties:
Maximum Bending Stress: Applying the flexure formula
Ans.
Ans.(s)max 4(103)(12)(3.40)
91.73 1779.07 psi 1.78 ksi
(s)max 4(103)(12)(10.5 3.40)
91.73 3715.12 psi 3.72 ksi
smax
91.73 in4
1
12 (0.5) 103 0.5(10)(5.5 3.40)2
2 1
12 (0.5)(33) 0.5(3)(3.40 2)2
1
12
(4) 0.53 4(0.5)(3.40 0.25)2
0.25(4)(0.5) 2[2(3)(0.5)] 5.5(10)(0.5)
4(0.5) 2[(3)(0.5)] 10(0.5) 3.40 in.
Determine the maximum tensile and compressivebending stress in the beam if it is subjected to a moment of 4 kip ft.
3 in.
D
AB
0.5 in.
M
0.5 in.
3 in.
C
10 in.
0.5 in.0.5 in. 75 mm
D
AB
12 mm
M
12 mm
75 mm
C
250 mm
12 mm12 mm
M = 6 kN · m.
(6)(99)(12) + 2[49.5(75)(12)] + 137(250)(12)
99(12) + 2[(75)(12)] + 250(12) 84.7 mm
I NA 1
12
(99)(123) + 99(12)(84.7 – 6)2
+ 2 1
12 (12)(753) + (12)(75)(84.7 – 49.5)2
+1
12 (12)(2503) + (12)(250)(137 – 84.7)2
34.277(106) mm4
6(106)(262 – 84.7)
34.277(106) 31.0 MPa
6(106)(4)(84.7)
34.277(106) 14.8 MPa
11–49.
1095
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 49/135
Assume failure due to tensile stress:
Assume failure due to compressive stress:
Ans.M = 3.27 kN # . = 2.50 kN # (controls)
1 5 = M( .6 - . 7)
smax = Mc
I ;
M = # . = 7.33 kN #
= M( . 7)
smax = My
I ;
I =
1
36(100) ( )3
= .8(10 )mm4
y(From base ) =
1
32 2
-2 = . 7 .28 86 mm50100
1
2886154
9.6 kN m
68 28 860
m
m
m
62 2 - 250100
.8(10 )mm41 6
.8(10 )mm41 6
11–50. A member has the triangular cross section shown.Determine the largest internal moment M that can beapplied to the cross section without exceeding allowabletensile and compressive stresses of (sallow)t = 154 MPa and(sallow)c = 105 MPa, respectively.
50 mm
50 mm
100 mm
M100 mm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1096
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 50/135
Ans.
Ans.(smax)c = Mc
I =
(12)( . 3)=
(smax)t =
My
I =
. 7)=
y =
1
3( .6) = . 7 .
c =
2
3( .6) = . 3 .
Ix =
1
36(100)(86.6)3
=
h = = .6 .86 mm
57 7
86 28 86 mm
86
28 86(12)(16 MPa
57 732 MPa
mm
2 2 - 250100
.8(10 )mm41 6
106
.8(10 )1 6
106
.8(10 )1 6
11–51. A member has the triangular cross section shown. If amoment of M = 1000 N · m is applied to the cross section,determine the maximum tensile and compressive bendingstresses in the member. Also, sketch a three-dimensional viewof the stress distribution action over the cross section.
50 mm
50 mm
100 mm
M100 mm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1097
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 51/135
*11–52. If the beam is subjected to an internal moment ofdetermine the maximum bending stress in
the beam. The beam is made from A992 steel. Sketch thebending stress distribution on the cross section.
M = 30 kN # m,
Section Properties: The moment of inertia of the cross section about the neutralaxis is
Maximum Bending Stress: The maximum bending stress occurs at the top andbottom surfaces of the beam since they are located at the furthest distance from theneutral axis.Thus, c = 75 mm = 0.075 m.
Ans.
At
The bending stress distribution across the cross section is shown in Fig.a.
s|y = 0.06 m = My
I =
30(103)(0.06)
15.165(10- 6)= 119 MPa
y = 60 mm = 0.06 m,
smax = Mc
I =
30(103)(0.075)
15.165(10 - 6)= 148 MPa
I =1
12 (0.1)(0.153) -
1
12 (0.09)(0.123) = 15.165(10 - 6) m4
50 mm
150 mm
15 mm
10 mm
15 mm
A
50 mm
M
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1098
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 52/135
11–53. If the beam is subjected to an internal moment ofdetermine the resultant force caused by the
bending stress distribution acting on the top flange A.M = 30 kN # m,
Section Properties: The moment of inertia of the cross section about the neutralaxis is
Bending Stress: The distance from the neutral axis to the top and bottom surfaces of flange A is yt = 75 mm = 0.075 m and yb = 60 mm = 0.06 m.
Resultant Force: The resultant force acting on flange A is equal to the volume of thetrapezoidal stress block shown in Fig.a. Thus,
Ans.= 200 296.74 N = 200 kN
FR =1
2 (148.37 + 118.69)(106)(0.1)(0.015)
sb = Myb
I =
30(103)(0.06)
15.165(10 - 6) = 118.69 = 119 MPa
st = Myt
I =
30(103)(0.075)
15.165(10 - 6) = 148.37 = 148 MPa
I =1
12 (0.1)(0.153) -
1
12 (0.09)(0.123) = 15.165(10- 6) m4
50 mm
150 mm
15 mm
10 mm
15 mm
A
50 mm
M
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1099
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 53/135
11–54. If the built-up beam is subjected to an internalmoment of determine the maximum tensileand compressive stress acting in the beam.
M = 75 kN # m,
300 mm
A
M
20 mm
10 mm
10 mm
150 mm
150 mm
150 mm
Section Properties: The neutral axis passes through centroid C of the cross sectionas shown in Fig. a. The location of C is
Thus, the moment of inertia of the cross section about the neutral axis is
Maximum Bending Stress: The maximum compressive and tensile stress occurs atthe top and bottom-most fiber of the cross section.
Ans.
Ans.(smax)t = Mc
I =
75(103)(0.2035)
92.6509(10 - 6)= 165 MPa
(smax)c = My
I =
75(103)(0.3 - 0.2035)
92.6509(10 - 6)= 78.1 MPa
= 92.6509(10 - 6) m 4
+ 2 c 1
12 (0.14)(0.013) + 0.14(0.01)(0.295 - 0.2035)2 d
+ 2 c 1
12 (0.01)(0.153) + 0.01(0.15)(0.225 - 0.2035)2 d
=1
12 (0.02)(0.33) + 0.02(0.3)(0.2035 - 0.15)2
I = ©I + Ad2
y = © y~A
©A
=0.15(0.3)(0.02) + 230.225(0.15)(0.01)] + 230.295(0.01)(0.14)4
0.3(0.02) + 2(0.15)(0.01) + 2(0.01)(0.14)
= 0.2035 m
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1100
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 54/135
11–55. If the built-up beam is subjected to an internal moment ofdetermine the amount of this internal moment
resisted by plate A.M = 75 kN # m,
300 mm
A
M
20 mm
10 mm
10 mm
150 mm
150 mm
150 mm
Section Properties: The neutral axis passes through centroid C of the cross sectionas shown in Fig. a. The location of C is
Thus, the moment of inertia of the cross section about the neutral axis is
= 92.6509(10 - 6) m4
+ 2 c1
12(0.14)(0.01
3
) + 0.14(0.01)(0.295 - 0.2035)
2
d
+ 2 c 1
12(0.01)(0.153) + 0.01(0.15)(0.225 - 0.2035)2 d
=1
12 (0.02)(0.33) + 0.02(0.3)(0.2035 - 0.15)2
I = I + Ad2
y = © y~A
©A =
0.15(0.3)(0.02) + 230.225(0.15)(0.01)] + 230.295(0.01)(0.14)40.3(0.02) + 2(0.15)(0.01) + 2(0.01)(0.14)
= 0.2035 m
Bending Stress: The distance from the neutral axis to the top andbottom of plate A is and
The bending stress distribution across the cross section of plate A is shown in Fig. b. The resultant forces of the tensileand compressive triangular stress blocks are
Thus, the amount of internal moment resisted by plate A is
Ans.= 50315.65 N # m = 50.3 kN # mM = 335144.46 c
2
3 (0.2035) d + 75421.50 c2
3 (0.0965) d
(FR)c =1
2(78.14)(106)(0.0965)(0.02) = 75 421.50 N
(FR)t =1
2(164.71)(106)(0.2035)(0.02) = 335 144.46 N
sb = Myb
I =
75(103)(0.2035)
92.6509(10 - 6)= 164.71 MPa (T)
st = Myt
I =
75(103)(0.0965)
92.6509(10 - 6)= 78.14 MPa (C)
yb = 0.2035 m.yt = 0.3 - 0.2035 = 0.0965 m
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1101
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 55/135
Section Property:
Bending Stress: Applying the flexure formula
Ans.
Ans.sB =8(103)(0.01)
17.8133(10 - 6)= 4.49 MPa (T)
sA =8(103)(0.11)
17.8133(10- 6)= 49.4 MPa (C)
s = My
I
I =1
12 (0.02)(0.223) +
1
12 (0.1)(0.023) = 17.8133(10 - 6) m4
*11–56. The aluminum strut has a cross-sectional area in theform of a cross. If it is subjected to the moment
determine the bending stress acting at points A and B, and show the results acting on volume elementslocated at these points.
M = 8 kN # m,
A
20 mm
B
20 mm
100 mm
50 mm50 mm
100 mm
M 8 kNm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1102
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 56/135
Section Property:
Bending Stress: Applying the flexure formula and ,
Ans.
sy= 0.01m =8(103)(0.01)
17.8133(10 - 6)= 4.49 MPa
smax =8(103)(0.11)
17.8133(10 - 6)= 49.4 MPa
s = My
Ismax =
Mc
I
I =1
12 (0.02)(0.223) +
1
12 (0.1)(0.023) = 17.8133(10 - 6) m4
11–57. The aluminum strut has a cross-sectional area in theform of a cross. If it is subjected to the moment
determine the maximum bending stress inthe beam, and sketch a three-dimensional view of the stressdistribution acting over the entire cross-sectional area.
M = 8 kN # m,
A
20 mm
B20 mm
100 mm
50 mm50 mm
100 mm
M 8 kNm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1103
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 57/135
Ans.
Ans.sC = My
I =
75(0.0175 - 0.01)
0.3633(10 - 6)= 1.55 MPa
sB = Mc
I =
75(0.0175)
0.3633(10 - 6)= 3.61 MPa
+ 2 c 1
12 (0.01)(0.043) + 0.01(0.04)(0.01252)d = 0.3633(10 - 5) m4
I =
1
12
(0.08)(0.01
3
) +
0.08(0.01)(0.0125
2
)
y =0.005(0.08)(0.01) + 230.03(0.04)(0.01)4
0.08(0.01) + 2(0.04)(0.01) = 0.0175 m
11–58. The aluminum machine part is subjected to a momentof Determine the bending stress created atpoints B and C on the cross section. Sketch the results on avolume element located at each of these points.
M = 75 kN # m.
M 75 Nm40 mm
10 mm
10 mm10 mm
20 mm
20 mm10 mm
10 mm
B A
N
C
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1104
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 58/135
Ans.
Ans.(smax)c = My
I =
75(0.0175)
0.3633(10 - 6)= 3.61 MPa
(smax)t = Mc
I =
75(0.050 - 0.0175)
0.3633(10 - 6)= 6.71 MPa
+ 2 c 1
12 (0.01)(0.043) + 0.01(0.04)(0.01252)d = 0.3633(10 - 5) m4
I =1
12 (0.08)(0.013) + 0.08(0.01)(0.01252)
y =0.005(0.08)(0.01) + 230.03(0.04)(0.01)4
0.08(0.01) + 2(0.04)(0.01) = 0.0175 m
11–59. The aluminum machine part is subjected to a momentof . Determine the maximum tensile andcompressive bending stresses in the part.
M = 75 kN # m
M 75 Nm40 mm
10 mm
10 mm10 mm
20 mm
20 mm10 mm
10 mm
B A
N C
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1105
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 59/135
Using flexure formula
Ans.
Ans.
Ans.FB =1
2( .5 + )( )( ) = . k
FA =1
2(22 + = 7 . 7 k
smax =
(250 -
= 4.9995 = 5. (Max)
sC =( -
= .1
sB = = .
sA = - )
=
s = My
I
= 4
+1
12 (75)( 3 + - 2
I =1
12 (125)( 3 + - 2 +
1
12 (25)( 3 + - 2
y = © y~A
©A =
+ +
( 5) + ( ) + ( ) = .
125(200)(25)12(25)(125) 237(75)(25)111 mm
75 25200 2525 12
200(25)(125 111)200)25 ) 125(25)(111 12)
25 ) 75(25)(237 111)
78.3(10 ) mm
111 25
20(10 )(111)
22 MPa78.3(10 )
20(10 )(
28 35 MPa
111)29 2 MPa
111)
3 5 MPa
8 6 N28.35)(25)(125)
29.1235 60 58 N25 75
MPa
6
6
6
6
78.3(10 )6
20(10 )6 25
78.3(10 )6
78.3(10 )6
20(10 )6
*11–60. The beam is subjected to a moment of 20 kN · m.
Determine the resultant force the bending stress produces
on the top flange A and bottom flange B. Also compute the
maximum bending stress developed in the beam.
75 mm
125 mm
25 mm
25 mm
200 mm
M 20 kN.m
25 mm
A
B
D
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1106
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 60/135
Using flexure formula
% of moment carried by web Ans.= * 100 = 22.6 %
= 40.623 kip # in. = #M = . ( ) + .5(7 )
FT =1
2 ( .1 )( 4)( ) = 5 k
FC =1
2 ( )( )( ) = . k
sB =(225 -
= .1
sA =(111 -
=
s =
My
I
=
+1
12 (75)( 3
+ -2
I =1
12 (125)( 3
+ -2
+1
12 (25)( 3
+ -2
y =© y~A
©A =
+ +
+ + = 111 mm
75(25)200(25)25(125)
237(75)(25)125(200)(25)12(25)(125)
200(25)(125 111)200 )125(25)(111 12)25 )
25 ) 75(25)(237 111)
25)22 MPa
111)29 2 MPa
23 65 N86 2522
41. N251129 2
65823 65 414.52 kN m
4.52
20
78.3(10 )6
20(10 )6
78.3(10 )6
78.3(10 )6
20(10 )6
11–61. The beam is subjected to a moment of 20 kN · m.
Determine the percentage of this moment that is resisted by
the web D of the beam.
75 mm
125 mm
25 mm
25 mm
200 mm
M 20 kN.m
25 mm
A
B
D
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1107
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 61/135
The moment of inertia of the cross-section about the neutral axis is
.
For point A, .
Ans.
For point B, .
Ans.
The state of stress at point A and B are represented by the volume element shownin Figs. a and b, respectively.
sB = MyB
I =
10(103)(0.125)
0.2417(10 - 3) = 5.172(106) Pa = 5.17 MPa (T)
yB = 0.125 m
sA = MyA
I =
10(103) (0.15)
0.2417(10 - 3) = 6.207(106) Pa = 6.21 MPa (C)
yA = C = 0.15 m
I =1
12 (0.2)(0.33) -
1
12 (0.16)(0.253) = 0.2417(10 - 3) m4
11–62. A box beam is constructed from four pieces of wood,glued together as shown. If the moment acting on the crosssection is , determine the stress at points A and Band show the results acting on volume elements located atthese points.
10 kN # m
20 mm 20 mm
250 mm
M 10 kNm
160 mm
25 mm
25 mm B
A
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1108
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 62/135
Ans.
Ans.
sC = My
I = = (T)
sB = My
I =
-= (T)
sA =My
I =
(40000)(12)( -= (C)
I =1
12 (50)(100 ) + (100)(50)(50 - 35) 2
- a 1
36 (50)(75)3
+ 1
2(50)(75)(75 - 35)2b =
4
y =-
12)(50)(75)
-12(50)(75)
=50(100)(50) 75(
35 mm100(50)
1.7(10 ) mm4
100 35)
40000(12)(35
1.53 MPa
25)0.235 MPa
40000(12)( )0.823 MPa
35
6
1.7(10 )6
1.7(10 )6
1.7(10 )6
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
11–63. The beam is subjected to a moment ofM = 40 N· m.Determine the bending stress acting at point A and B. Also,stetch a three-dimensional view of the stress distributionacting over the entire cross-sectional area. 75 mm
25 mm
25 mm
A
B
M 40 Nm
1109
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 63/135
Ans.smax = Mc
I =
14 p(68.75)4
=22500(68.75)
88 MPa
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
*11–64. The axle of the freight car is subjected to wheelloading of 90 kN. If it is supported by two journal bearings atC and D, determine the maximum bending stress developedat the center of the axle, where the diameter is 140 mm.
C D A B
90 kN 90 kN250 mm 250 mm
1500 mm
1110
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 64/135
(a)
Ans.
(b)
Ans.smax =
50(0.04)
4.021238(10-6)= 497 kPa
= 4.021238(10-6) m4
=
(0.04)4
2 sin-1
a y
0.04b ̀0.04
-0.04
= 4 c (0.04)4
8 sin-1a y
0.04b -
1
8y2 (0.04)2
- y2(0.042- 2y2) d ̀
-0.0
0.04
4
2L0.04
-0.04
y2zdy = 4L
0.04
-0.04
y22 (0.04)
2- y
2 dy
z = 2 0.0064 - 4y2= 22 (0.04)2
- y2
=
smax
c Ly22zdy
M =
s
maxc LA
y2dA
smax =
Mc
I =
50(0.04)
4.021238(10-6)= 497 kPa
I =
1
4 p ab3
=
1
4 p(0.08)(0.04)3
= 4.021238(10-6) m4
A shaft is made of a polymer having an ellipticalcross-section. If it resists an internal moment of
50 N m, determine the maximum bending stressdeveloped in the material (a) using the flexure formula,
where (b) using integration.
Sketch a three-dimensional view of the stress distributionacting over the cross-sectional area.
lz =14p(0.08 m)(0.04 m)3,
#M =
y
z x
M 50 Nm
80 mm
160 mm
y———(40)
2
2
z———(80)
2
2 1
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
11–65.
1111
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 65/135
(a)
Ans.
(b)
Ans.smax = 249 kPa
50 = 201.06(10-6)smax
50 = 2asmax
0.04 bL
0.08
0
z2a1 -
z2
(0.08)2b1>2
(0.04)dz
M =
LA
z(s dA) =
LA
zasmax
0.08
b(z)(2y)dz
smax =
Mc
I =
50(0.08)
16.085(10-6)= 249 kPa
I =
1
4 p ab3
=
1
4 p(0.04)(0.08)3
= 16.085(10-6) m4
Solve Prob. 6–65 if the moment isapplied about the y axis instead of the x axis. HereIy =
14 p(0.04 m)(0.08 m)3.
M = 50 kN m# y
z x
M 50 Nm
80 mm
160 mm
y———(40)
2
2
z———(80)
2
2 1
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
11–66.
1112
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 66/135
Absolute Maximum Bending Stress: The maximum moment isas indicated on the moment diagram.Applying the flexure formula
Ans.= 158 MPa
=11.34(103)(0.045)
p
4 (0.0454)
smax = Mmax c
I
Mmax = 11.34 kN # m
11–67. The shaft is supported by smooth journal bearings at A and B that only exert vertical reactions on the shaft. Ifd = 90 mm, determine the absolute maximum bending stressin the beam, and sketch the stress distribution acting overthe cross section. B
d
A
3 m 1.5 m
12 kN/ m
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1113
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 67/135
Allowable Bending Stress: The maximum moment is asindicated on the moment diagram.Applying the flexure formula
Ans.d = 0.08626 m = 86.3 mm
180A106 B =11.34(103) Ad2 B
p
4 Ad2 B4
smax = sallow = Mmax c
I
Mmax = 11.34 kN # m
*11–68. The shaft is supported by smooth journal bearings at A and B that only exert vertical reactions on the shaft.Determine its smallest diameter d if the allowable bendingstress is .sallow = 180 MPa
B
d
A
3 m 1.5 m
12 kN/ m
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1114
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 68/135
Section Property:
For section (a)
For section (b)
Maximum Bending Stress: Applying the flexure formula
For section (a)
For section (b)
Ans.smin =150(103)(0.18)
0.36135(10 - 3) = 74.72 MPa = 74.7 MPa
smax =150(103)(0.165)
0.21645(10 - 3) = 114.3 MPa
smax = Mc
I
I =1
12 (0.2) A0.363 B -
1
12 (0.185) A0.33 B = 0.36135(10 - 3) m4
I =1
12 (0.2) A0.333 B -
1
12 (0.17)(0.3)3 = 0.21645(10 - 3) m4
11–69. Two designs for a beam are to be considered.Determine which one will support a moment of
with the least amount of bending stress.What is that stress?M = 150 kN # m
200 mm
300 mm
(a) (b)
15 mm
30 mm
15 mm
200 mm
300 mm
30 mm
15 mm
30 mm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1115
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 69/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
Ans. 22.1 ksi
smax
27 000(4.6091 0.25)
5.9271
max 300(9 1.5)(12) 27 000 lb in.
0.19635(4.6091)2 5.9271 in4
14
p(0.5)4 1
4 p(0.3125)4 0.4786(6.50 4.6091)2
1
4p(0.25)4
0 (6.50)(0.4786)
0.4786 0.19635 4.6091 in.
11–70. The simply supported truss is subjected to the centraldistributed load. Neglect the effect of the diagonal lacing anddetermine the absolute maximum bending stress in the truss.The top member is a pipe having an outer diameter of 1 in.and thickness of and the bottom member is a solid rodhaving a diameter of 1
2 in.
316 in.,
6 ft
5.75 in.
6 ft 6 ft
100 lb ft
The top member is a pipe having an outer diameter of25 mm and thickness of 5 mm, and the bottom member is asolid rod having a diameter of 12 mm.
1.35 kN
2.7 m
1.35 kN
0.45
0 + (160)(314.16)
314.16 + 113.10 117.65 mm
I 1
4 (12.5)4 –
1
4 (7.5)4 + 314.16(160 – 117.65)2 +
1
4 (6)4
+ 113.10(117.65)2 = 2.1466(106) mm4
1.35(2.7) – 0.5(1.5)(0.92) = 3.0375 kN · m
3.0375(106)(117.65 + 6)
2.1466(106)
175.0 MPa
1.8 m
141.5 mm
1.8 m 1.8 m
1.5 kN/m
1116
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 70/135
Boat:
a
Assembly:
a
Ans.smax = Mc
I =
(12)(37.5)=
I =1
12 (43.75)(75)3 -
1
12 (37.5)(43.75) 3 = 4
Cy = 0
Cy + 0 0 - 00 = 0+ c ©F y = 0;
ND = 0 0
-ND( ) + 00( ) = 0+ ©MC = 0;
By = .
. - + By = 0+ c ©F y = 0;
NA = .
-NA(2.7) + = 0+ ©MB = 0;
Bx = 0:+ ©Fx = 0;
10000(1.5)
5555 55 N
5555 55 10000
4444 45 N
3 100 2.7
9 0 N
1009 0
100 N
1.28(10 ) mm
146.5 MPa
6
(10 )6
1.28(10 )6
5
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
11–71. The boat has a weight of 10000 N and a center ofgravity atG. If it rests on the trailer at the smooth contact A and can be considered pinned at B, determine the absolutemaximum bending stress developed in the main strut of thetrailer. Consider the strut to be a box-beam having thedimensions shown and pinned at C .
0.3 m
A
B
C
0.9 m0.3 m
1.5 m1.2 m
D
G
45 mm
75 mm 45 mm
40 mm
1117
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 71/135
Ans.s = Mc
I =
1
4 p(18.75)4
=
Mmax = 506.25 N-m.
506250(18.75)98 MPa
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
*11–72. Determine the absolute maximum bending stressin the 40-mm-diameter shaft which is subjected to theconcentrated forces. The sleeve bearings at A and B supportonly vertical forces.
300 mm
450 mm
B
A
1800 N
375 mm
1350 N
1118
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 72/135
Ans.d = .
c = 6 .
s = Mc
I; 154(10 )3 =
c
1
4 pc4
Mmax =
506250
1 mm
32 mm
506.25 N-m.
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
11–73. Determine the smallest allowable diameter of theshaft which is subjected to the concentrated forces. Thesleeve bearings at A and B support only vertical forces, andthe allowable bending stress is sallow = 154 MPa.
300 mm
450 mm
B
A
1800 N
375 mm
1350 N
1119
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 73/135
Ans.= 5
smax = Mc
I =
)
I = 14
p(5 )4 = 4
M = = #
w1 = > .w1( ) = 00;
w2 = > .1
2 w2( ) = 1800; 144 N mm25
3640 90 N mm
31860 N mm1800(17.7)
491 mm
531860 (
491324.4 MPa
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
11–74. The pin is used to connect the three links together.Due to wear, the load is distributed over the top and bottomof the pin as shown on the free-body diagram. If thediameter of the pin is 10 mm, determine the maximumbending stress on the cross-sectional area at the centersection a–a. For the solution it is first necessary to determinethe load intensitiesw1 and w2.
3600 N
1800 N 1800 N
25 mm
10 mm
40 mm
25 mm
a
a
w2 w2
w1
1120
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 74/135
Shear and Moment Diagrams:As shown in Fig. a.
Maximum Moment: Due to symmetry, the maximum moment occurs in region BC of the shaft. Referring to the free-body diagram of the segment shown in Fig.b.
Section Properties: The moment of inertia of the cross section about the neutralaxis is
Absolute Maximum Bending Stress:
Ans.smax = Mmaxc
I =
2.25 A103 B(0.04)
1.7038 A10- 6 B = 52.8 MPa
I = p
4 A0.044 - 0.0254 B = 1.7038 A10- 6 B m4
11–75. The shaft is supported by a smooth thrust bearing at A and smooth journal bearing atD. If the shaft has the crosssection shown, determine the absolute maximum bendingstress in the shaft.
AC
DB
3 kN 3 kN
0.75 m 0.75 m1.5 m
40 mm25 mm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1121
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 75/135
The moment of inertia of the cross section about the neutral axis is
Thus,
Ans.
The bending stress distribution over the cross section is shown in Fig.a.
M = 195.96 (103) N # m = 196 kN # m
smax = Mc
I; 80(106) =
M(0.15)
0.36742(10 - 3)
I =1
12 (0.3)(0.33) -
1
12 (0.21)(0.263) = 0.36742(10 - 3) m4
*11–76. Determine the moment M that must be applied tothe beam in order to create a maximum stress of 80 MPa.Also sketch the stress distribution acting over the crosssection.
260 mm
20 mm30 mm
300 mm
M
30 mm
30 mm
20 mm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1122
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 76/135
Section Properties: The neutral axis passes through centroid C of the cross section
as shown in Fig. a. The location ofC is
Thus, the moment of inertia of the cross section about the neutral axis is
y = ©y~A
©A =
2[37.5(75)(25)] + +
2( )( ) + ( ) + ( ) = .
=
+ 1
12
(25)( 3+ (25)(100)(150 - 85)2
= 2 c 1
12(25)( 3
+ ( )( - .5) 2d + 1
12(200)( 3
+ 200(25)(87.5 + 85)
I = ©I + Ad2
Maximum Bending Stress: The maximum compressive and tensile stress occurs atthe top and bottom-most fibers of the cross section.
Ans.
Ans.
The bending stresses at y y
The bending stress distribution across the cross section is shown inFig. b.
s|y= - 0.3889 in. =My
I =
(9.7)= .1 (T)
s|y= 0.6111 in. =My
I = = .7 (C)
= -=
(smax)t =My
I =
)= .
(smax)c = Mc
I =
8 )= 13 MPa
150(100)(25)87.5(200)(25)85 mm
200 25 100 2575 25
25 ) 275 ) 25 75 85 37
100 )
23.155(10 ) mm6
200 53(10 )(12) (
23.155
59 MPa
15.3 mm and 9.7 are
5
mm
1 MPa
1 MPa
( ) ( )
(12)(8
(12)(15.3)
4
6-
(10 )6
3(10 )6
23.155(10 )6
3(10 )6
3(10 )(12)6
23.155(10 )6
23.155(10 )6
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
11–77. If the beam is subjected to an internal moment ofM = 3 kN # m, determine the maximum tensile andcompressive stress in the beam. Also, sketch the bendingstress distribution on the cross section. 100 mm
25 mm
75 mm
75 mm
75 mm
25 mm
25 mm
25 mm
A
M
1123
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 77/135
Section Properties: The neutral axis passes through the centroid C of the crosssection as shown in Fig. a. The location ofC is given by
Thus, the moment of inertia of the cross section about the neutral axis is
y = ©y~A
©A =
2[37.5(75) 25) + +
2(75)(25) + + = 8 .
=
+1
12
(25)(
3+
(25)(100)(150 -
85)
2
+1
12 (200)(253) + 85)= 2c 1
12 (25)( 3
+ - .5)2 dI = ©I + Ad2
Allowable Bending Stress: The maximum compressive and tensile stress occurs atthe top and bottom-most fibers of the cross section.
For the bottom-most fiber,
Ans.M = (controls)
=M(85)
(sallow)t =My
I;
M =
=M(200 -
(sallow)c = Mc
I;
150(100)(25)]87.5(200)(25)5 mm
200(25) 100(25)
200(25)(87.53725(75)(8575 )
100 )
2
85)21
4.3 kN-m
14
4 kN-m
(
23.155(10 )6 mm4
23.155(10 )6
23.155(10 )6
+
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
11–78. If the allowable tensile and compressive stress forthe beam are (sallow)t = 14 MPa and (sallow)c = 21 MPa,respectively, determine the maximum allowable internalmoment M that can be applied on the cross section.
100 mm
25 mm
75 mm
75 mm
75 mm
25 mm
25 mm
25 mm
A
M
1124
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 78/135
Section Properties: The neutral axis passes through the centroid C of the crosssection as shown in Fig. a. The location ofC is given by
Thus, the moment of inertia of the cross section about the neutral axis is
y = ©~yA
©A =
2[37.5(75)(25) + +
2(75)(25) + + =
=
+1
12(25)( 3
+ (25)(100)(150 -2
= 2 c 1
12(25)( 3
+ -2d +
1
12(200)( 3
+
I = © I + Ad 2
Bending Stress: The distance from the neutral axis to the top and bottom of board Ais yt = 200 85 = 115 mm and yb = 100 85 = .We have
Resultant Force: The resultant force acting on board A is equal to the volume of thetrapezoidal stress block shown in Fig.b. Thus,
Ans.FR =1
2(1.3 + =
sb =
Myb
I =
(15)
= .7
st =Myt
I =
(115)= 1
--
150(100)(25)]87.5(200)(25)85 mm
100(25)200(25)
75 ) 25(75)(85 37.5) 25 ) 200(25)(87.5 85)
100 ) 85)
+2
15 mm
3 MPa
1 MPa
19 kN1.7)(25)(100)
23.155(10 )6 mm4
23.155(10 )6
23.155(10 )6
(10 )(12)63
(10 )(12)63
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
11–79. If the beam is subjected to an internal moment ofM = 3 kN # m, determine the resultant force of the bendingstress distribution acting on the top vertical board A. 100 mm
25 mm
75 mm
75 mm
75 mm
25 mm
25 mm
25 mm
A
M
1125
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 79/135
*11–80. If the beam is subjected to an internal moment ofdetermine the bending stress developed
at points A,B, and C . Sketch the bending stress distributionon the cross section.
M = 100 kN # m,
M
300 mm
150 mm
30 mm
150 mm
C
30 mm
B
A
Section Properties: The neutral axis passes through centroid C of the cross sectionas shown in Fig. a. The location of C is
Thus, the moment of inertia of the cross section about the neutral axis is
y = © ~yA
©A =
0.015(0.03)(0.3) + 0.18(0.3)(0.03)
0.03(0.3) + 0.3(0.03) = 0.0975 m
= 0.1907(10 - 3) m4
+ 0.03(0.3)(0.18 - 0.0975)2
I =1
12 (0.3)(0.033) + 0.3(0.03)(0.0975 - 0.015)2 +
1
12(0.03)(0.33)
Bending Stress: The distance from the neutral axis to points A, B, and C is
Ans.
Ans.
Ans.
Using these results, the bending stress distribution across the cross section is shownin Fig. b.
sC = MyC
I =
100(103)(0.0675)
0.1907(10 - 3) = 35.4 MPa (T)
sB = MyB
I =
100(103)(0.0975)
0.1907(10 - 3) = 51.1 MPa (T)
sA = MyA
I =
100(103)(0.2325)
0.1907(10 - 3) = 122 MPa (C)
0.0675 m.yC = 0.0975 - 0.03 =yB = 0.0975 m, and0.33 - 0.0975 = 0.2325 m,yA =
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1126
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 80/135
11–81. If the beam is made of material having an allowabletensile and compressive stress of and
, respectively, determine the maximumallowable internal moment M that can be applied to thebeam.
(sallow)c = 150 MPa(sallow)t = 125 MPa
M
300 mm
150 mm
30 mm
150 mm
C
30 mm
B
A
Section Properties: The neutral axis passes through centroid C of the cross sectionas shown in Fig. a.The location of C is
Thus, the moment of inertia of the cross section about the neutral axis is
=0.015(0.03)(0.3) + 0.18(0.3)(0.03)
0.03(0.3) + 0.3(0.03) = 0.0975 m y =
© ~yA
©A
= 0.1907(10 - 3) m4
+ 0.03(0.3)(0.18 - 0.0975)2
I =1
12 (0.3)(0.033) + 0.3(0.03)(0.0975 - 0.015)2 +
1
12 (0.03)(0.33)
Allowable Bending Stress: The maximum compressive and tensile stress occurs atthe top and bottom-most fibers of the cross section. For the top-most fiber,
Ans.
For the bottom-most fiber,
M = 244 471.15 N # m = 244 kN # m
125(106) = M(0.0975)
0.1907(10 - 3)(sallow)t =
My
I
M = 123024.19 N # m = 123 kN # m (controls)
150(106) = M(0.33 - 0.0975)
0.1907(10 - 3)(sallow)c =
Mc
I
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1127
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 81/135
Support Reactions: Shown on the free-body diagram of the shaft, Fig.a,
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c. Asindicated on the moment diagram, the maximum moment is .
Section Properties: The moment of inertia of the cross section about the neutralaxis is
Absolute Maximum Bending Moment: Here,
smax = Mmaxc
I = = 2
c =3
2
= .
I =1
4 p(37 )4 = 4
|Mmax| = 8000 N-m
1.472 mm
37 mm
8(10 )(12)(37)00 MPa
(10 )6
6
1.472 (10 )6
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
11–82. The shaft is supported by a smooth thrust bearing at A and smooth journal bearing atC . If d = 75mm, determinethe absolute maximum bending stress in the shaft. A C D
d
B
1 m 1 m
16000 N
8000 N1 m
1128
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 82/135
Support Reactions: Shown on the free-body diagram of the shaft, Fig.a.
Maximum Moment: As indicated on the moment diagram, Figs. b and c, themaximum moment is
Section Properties: The moment of inertia of the cross section about the neutralaxis is
Absolute Maximum Bending Moment:
Use Ans.d = mm
d =
3 =
ad2b
p
64 d4
sallow = Mc
I;
I =1
4pad
2b 4
= p
64 d4
|Mmax| = 8000 N-m.
8000000(12)
168(10 )
75 mm
75
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
11–83. The shaft is supported by a smooth thrust bearingat A and smooth journal bearing at C . If the material hasan allowable bending stress of sallow = 168 MPa, determinethe required minimum diameter d of the shaft to thenearest mm.
A C D
d
B
1 m 1 m
16000 N
8000 N1 m
1129
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 83/135
*11–84. If the intensity of the load w = 15 kN m, determinethe absolute maximum tensile and compressive stress in thebeam.
>
6 m
150 mm
300 mm
AB
w
Support Reactions: Shown on the free-body diagram of the beam, Fig.a.
Maximum Moment: The maximum moment occurs when . Referring to thefree-body diagram of the beam segment shown in Fig. b,
a
Section Properties: The moment of inertia of the cross section about the neutralaxis is
Absolute Maximum Bending Stress: The maximum compressive and tensile stressesoccur at the top and bottom-most fibers of the cross section.
Ans.
Ans.(smax)t = My
I =
67.5(103)(0.1)
0.1125(10 - 3) = 60 MPa (T)
(smax)c = Mc
I =
67.5(103)(0.2)
0.1125(10 - 3) = 120 MPa (C)
I =1
36 (0.15)(0.33) = 0.1125(10 - 3) m4
Mmax = 67.5 kN # mMmax + 15(3)a 3
2b - 45(3) = 0+gM = 0;
x = 3 m45 - 15x = 0+ cg Fy = 0;
V = 0
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1130
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 84/135
11–85. If the material of the beam has an allowable bendingstress of , determine the maximumallowable intensity w of the uniform distributed load.
sallow = 150 MPa
Support Reactions: As shown on the free-body diagram of the beam, Fig.a,
Maximum Moment: The maximum moment occurs when . Referring to thefree-body diagram of the beam segment shown in Fig.b,
a
Section Properties: The moment of inertia of the cross section about the neutralaxis is
Absolute Maximum Bending Stress: Here,
Ans.w = 18750 N>m = 18.75 kN>m
150(106) =
9
2 w(0.2)
0.1125(10 - 32 sallow = Mc
I;
c =2
3 (0.3) = 0.2 m.
I =1
36(0.15)(0.33) = 0.1125(10 - 3) m4
Mmax =9
2 wMmax + w(3)a 3
2b - 3w(3) = 0+gM = 0;
x = 3 m3w - wx = 0+ cgFy = 0;
V = 0
6 m
150 mm
300 mm
AB
w
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1131
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 85/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
11–86. Determine the absolute maximum bending stressin the 2-in.-diameter sh aft w hich i s subjected to theconcentrated forces. The journal bearings at A and B onlysupport vertical forces.
15 in.
15 in. B
A
800 lb
30 in.
600 lb
The FBD of the shaft is shown in Fig.a.
The shear and moment diagrams are shown in Fig.b and c, respectively. Asindicated on the moment diagram, .
The moment of inertia of the cross-section about the neutral axis is
Here, . Thus
Ans. 19.1 ksi
19.10(103) psi15000(1)
0.25 p
smax max
1 in
p
4 (14) 0.25 p 4
max 15000 lb in
4 kN 3 kN
375 mm 375 mm 750 mm
A y = 4.5 kN B y = 2.5 kN
M (kN · m)
V (kN)
4.5
0.5 1500
–2.5
375
x (mm) x (mm)
750375 1500
1.68751.875
375 mm
375 mm B
A
4 kN
750 mm
3 kNin the 50-mm-diameter shaft which is subjected to the
1.875 kN · m.
(254) = 306796 mm4
1.875(106)(25)
306796
152.8 MPa
750
1132
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 86/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
Determine the smallest allowable diameter of theshaft which is subjected to the concentrated forces. The
journal bearings at A and B only support vertical forces.The allowable bending stress is sallow 22 ksi.
15 in.
15 in. B
A
800 lb
30 in.
600 lb
The FBD of the shaft is shown in Fig.a
The shear and moment diagrams are shown in Fig.b and c respectively. Asindicated on the moment diagram,
The moment of inertia of the cross-section about the neutral axis is
Here, . Thus
Ans. 1.908 in 2 in.
sallow max
;
22(103) 15000(2)
p464
2
p
4
24
p
64 4
max 15,000 lb in
The allowable bending stress isallow = 150 MPa.
1.875 kN · m.
150 1.875(106)
d4/64
d 50.3 mm
375 mm
375 mm B
A
4 kN
750 mm
3 kN
4 kN 3 kN
375 mm 375 mm 750 mm
A y = 4.5 kN B y = 2.5 kN
M (kN · m)
V (kN)
4.5
0.5 1500
–2.5
375
x (mm) x (mm)
750375 1500
1.68751.875
750
11–87 .
1133
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 87/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
Absolute Maximum Bending Stress: The maximum moment isas indicated on moment diagram. Applying the flexure formula
Ans.smax max
44.8(12)(4.5)1
12 (9)(9)3 4.42 ksi
max
*6–88. If the beam has a square cross section of 9 in. oneach side, determine the absolute maximum bending stressin the beam.
A
B
8 ft 8 ft
800 lb/ ft1200 lb
43.5 M (kN · m)
V (kN)
6
8 16
–76.875
–15
8 16 x (m) x (m)
A
B
2.5 m 2.5 m
15 kN/m6 kN*11–88. If the beam has a square cross section of 225 mm on
76.875(106)(112.5)112(225)(2253)
= 40.49 MPa
76 875 # mkN
1134
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 88/135
Allowable Bending Stress: The maximum moment is asindicated on moment diagram.Applying the flexure formula
Ans.a = 0.06694 m = 66.9 mm
150A106 B =7.50(103) Aa2 B
112 a4
smax = sallow = Mmax c
I
Mmax = 7.50 kN # m
11–89. If the compound beam in Prob. 11–42 has a squarecross section of side length a, determine the minimum valueof a if the allowable bending stress is .sallow = 150 MPa
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1135
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 89/135
Absolute Maximum Bending Stress: The maximum moment is as
indicated on the moment diagram.Applying the flexure formula
Ans.smax = Mmax c
I =
23w0 L2
216 Ah2 B1
12 bh3 =
23w0 L2
36bh2
Mmax =23w0 L2
216
11–90. If the beam in Prob. 11–28 has a rectangular crosssection with a width b and a height h, determine theabsolute maximum bending stress in the beam.
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1136
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 90/135
The FBD of the shaft is shown in Fig. a
The shear and moment diagrams are shown in Fig. b and c, respectively. As
indicated on the moment diagram, .
The moment of inertia of the cross section about the neutral axis is
Here, . Thus
Ans.= 119 MPa = 119.37(106) Pa
smax = Mmax c
I =
6(103)(0.04)
0.64(10 - 6)p
c = 0.04 m
I = p
4 (0.044) = 0.64(10 - 6)p m4
Mmax = 6 kN # m
11–91. Determine the absolute maximum bending stress inthe 80-mm-diameter shaft which is subjected to theconcentrated forces. The journal bearings at A and B onlysupport vertical forces.
0.5 m 0.6 m0.4 m
20 kN
A B
12 kN
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1137
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 91/135
The FBD of the shaft is shown in Fig. a.
The shear and moment diagrams are shown in Fig. b and c, respectively.As indicated
on the moment diagram, .
The moment of inertia of the cross section about the neutral axis is
Here, . Thus
Ans.d = 0.07413 m = 74.13 mm = 75 mm
sallow =
Mmax c
I ;
150(106
) =
6(103)(d
>2)
pd4>64
c = d>2
I = p
4 ad
2b4
= pd4
64
Mmax = 6 kN # m
*11–92. Determine, to the nearest millimeter, the smallestallowable diameter of the shaft which is subjected to theconcentrated forces. The journal bearings at A and B onlysupport vertical forces. The allowable bending stress is
.sallow = 150 MPa0.5 m 0.6 m0.4 m
20 kN
A B
12 kN
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1138
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 92/135
Ans.
Note that 5. OK6 sY = 0
smax = = 5.smax = Mc
I;
5184000(37.5)17 135 MPa
1.11(10 )
17 135 MPa 42 MPa
6
++
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
11–93. The wing spar ABD of a light plane is made from2014–T6 aluminum and has a cross-sectional area of800 mm , a depth of 75 mm, and a moment of inertia about2
its neutral axis of 1.11(10-6) m4. Determine the absolutemaximum bending stress in the spar if the anticipated loadingis to be as shown. Assume A,B, andC are pins. Connection ismade along the central longitudinal axis of the spar.
0.6 m DB A
C
0.9 m 1.8 m
100 N/m
1139
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 93/135
Ans.P = 10.4 kN
10(106) =1.5P(0.125)
1.953125(10 - 4)
smax = Mc
I
I =1
12 (0.15)(0.253) = 1.953125(10 - 4) m4
11–94. The beam has a rectangular cross section as shown.Determine the largest load P that can be supported on itsoverhanging ends so that the bending stress does notexceed .smax = 10 MPa
P P
1.5 m 1.5 m 1.5 m
150 mm
250 mm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1140
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 94/135
Ans.smax = Mc
I =
18000(0.125)
1.953125(10 - 4) = 11.5 MPa
I =1
12 (0.15)(0.253) = 1.953125(10 - 4) m4
M = 1.5P = 1.5(12)(103) = 18000 N # m
11–95. The beam has the rectangular cross section shown. If P = 12 kN, determine the absolute maximum bending stressin the beam. Sketch the stress distribution acting over thecross section.
P P
1.5 m 1.5 m 1.5 m
150 mm
250 mm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1141
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 95/135
Thus, from the above equations,
Ans.
Ans.
Ans.P = 4 k
h =
b = .
b = .
(600)2 - 3b2 = dP
db = 0
b(600)2 - b3 = P
bh2 =6
(1200 P)
sallow =6 M max
bh2
sallow = Mc
I =
Mmax(h2)
112(b)(h)3
Mmax = P
2 =
(600)2 = b2 + h2
1200 P2400
56
128.57
128.57
346.41
346 41 mm
490 mm
6 7 N
mm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
*11–96. A log that is 0.6 m in diameter is to be cut into arectangular section for use as a simply supported beam. Ifthe allowable bending stress for the wood issallow = 56 MPa,determine the required width b and height h of the beamthat will support the largest load possible. What is this load?
2.4 m
0.6 m
h
b
2.4 m
P
1142
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 96/135
Ans.P = k
=P ( 2 )
112(200)(566)3
sallow = Mmax c
I
Mmax = P
2 (2400) = P
h = 6 .
2 = h2 + 2
600 200
56 mm
1200
5661200
56
500 N
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
11–97. A log that is 0.6 m in diameter is to be cut into arectangular section for use as a simply supported beam. Ifthe allowable bending stress for the wood is sallow = 56 MPa,determine the largest load P that can be supported if thewidth of the beam is b = 200 mm.
2.4 m
0.6 m
h
b
2.4 m
P
1143
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 97/135
Absolute Maximum Bending Stress: The maximum moment isas indicated on moment diagram.Applying the flexure formula
Ans.smax = Mmax c
I =
294(10 )(200)112 (200)(400 )3
= .
Mmax = 294 kN #
200 mm
400 mm
m
55 1 MPa
6
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
11–98. If the beam in Prob. 11–18 has a rectangular crosssection with a width of 200 mm and a height of 400 mm,determine the absolute maximum bending stress in the beam.
1144
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 98/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
The maximum moment occurs at the fixed support A. Referring to the FBD shownin Fig. a,
a
The moment of inertia of the about the neutral axis is .Thus,
Ans. 5600 psi 5.60 ksi
smax
16800(12)(3)
108
1
12 (6)(63) 108 in4
max 16800 lb ft
0;
max 400(6)(3) 1
2 (400)(6)(8) 0
6–99. If the beam has a square cross section of 6 in. oneach side, determine the absolute maximum bending stressin the beam.
A
B
6 ft 6 ft
400 lb/ ft11–99. If the beam has a square cross section of 150 mm on
6(1.8)(0.9) –1
2 (6)(1.8)(2.4) = 0
22.60 kN · m
22.68(106)(75)
42.1875(106)
40.3 MPa
A
B
1.8 m 1.8 m
6 kN/m
0.9 m
6(1.8) kN 12 (6)(1.8) kN
2.4 m(150)(1503) = 42.1875(106) mm4
1145
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 99/135
Support Reactions: Shown on the free-body diagram of the beam, Fig.a.
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c.As indicated on the moment diagram,
Section Properties: The moment of inertia of the cross section about the neutralaxis is
Absolute Maximum Bending Stress: Here,
Ans.smax =
M max c
I =
24(103)(0.225)
1.0477(10- 3)= 5.15 MPa
c =0.45
2 = 0.225 m.
= 1.0477(10- 3) m4
I =1
12 (0.175)(0.453) -
1
12 (0.125)(0.33)
Mmax = 24kN # m.
*11–100. If d = 450 mm, determine the absolute maximumbending stress in the overhanging beam.
4 m
8 kN/ m
2 m
12 kN
d
75 mm
25 mm
125 mm
25 mm
75 mm
A
B
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1146
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 100/135
11–101. If wood used for the beam has an allowablebending stress of determine the minimumdimension d of the beam’s cross-sectional area to thenearest mm.
sallow = 6 MPa,
Support Reactions: Shown on the free-body diagram of the beam, Fig.a.
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c.As indicated on the moment diagram,
Section Properties: The moment of inertia of the cross section about the neutralaxis is
Mmax = 24kN # m.
Absolute Maximum Bending Stress: Here, c = d
2.
= 4.1667(10- 3)d3+ 4.6875(10 - 3)d2
- 0.703125(10 - 3)d + 35.15625(10 - 6)
I =1
12 (0.175)d3
-1
12 (0.125)(d - 0.15)3
Solving,
Ans.d = 0.4094 m= 410 mm
4.1667(10- 3
)d3
+ 4.6875(10- 3
)d2
- 2.703125(10- 3
)d + 35.15625(10- 6
) = 0
6(106) =
24(103)d
2
4.1667(10 - 3)d3+ 4.6875(10- 3)d2
- 0.703125(10 - 3)d + 35.15625(10- 6)
sallow = Mc
I ;
4 m
8 kN/ m
2 m
12 kN
d
75 mm
25 mm
125 mm
25 mm
75 mm
A
B
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1147
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 101/135
11–102. If the concentrated force P = 2 kN is applied at thefree end of the overhanging beam, determine the absolutemaximum tensile and compressive stress developed inthe beam.
= 68.0457(10- 6) m 4
+ 0.15(0.025)(0.2125 - 0.13068)2+
1
12 (0.15)(0.0253)= 2 c 1
12 (0.025)(0.23) + 0.025(0.2)(0.13068 - 0.1)2 d
I = gI + Ad 2
Absolute Maximum Bending Stress: The maximum tensile and compressive stressoccurs at the top and bottom-most fibers of the cross section.
Ans.
Ans.(smax)c =M max c
I =
2(103)(0.13068)
68.0457(10 - 6)= 3.84 MPa
(smax )t =M max y
I =
2(103)(0.225 - 0.13068)
68.0457(10- 6)= 2.77 MPa
2 m 1 m
150 mm
200 mm
25 mm
25 mm 25 mm
B
A
P
Support Reactions: Shown on the free-body diagram of the beam, Fig.a.
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c.As indicated on the moment diagram, the maximum moment is
Section Properties: The neutral axis passes through the centroid C of the crosssection as shown in Fig. d.
The location of C is given by
Thus, the moment of inertia of the cross section about the neutral axis is
y = gy~A
gA =2[0.1(0.2)(0.025)] + 0.2125(0.025)(0.15)
2(0.2)(0.025) + 0.025(0.15) = 0.13068 m
ƒMmax ƒ = 2 kN # m.
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1148
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 102/135
11–103. If the overhanging beam is made of wood havingthe allowable tensile and compressive stresses of
and determine themaximum concentrated force P that can applied at thefree end.
(sallow )c = 5 MPa,(sallow )t = 4 MPa
Support Reactions: Shown on the free-body diagram of the beam, Fig.a.
Maximum Moment: The shear and moment diagrams are shown in Figs. b and c.As indicated on the moment diagram, the maximum moment is
Section Properties: The neutral axis passes through the centroid C of the crosssection as shown in Fig.d.The location of C is given by
ƒMmax ƒ = P .
The moment of inertia of the cross section about the neutral axis is
= 68.0457(10- 6) m4
+ 0.15(0.025)(0.2125 - 0.13068)2+
1
12 (0.15)(0.0253)= 2 c 1
12 (0.025)(0.23) + 0.025(0.2)(0.13068 - 0.1)2 d
I = gI + Ad2
y =
g y~AgA =
2[0.1(0.2)(0.025)] + 0.2125(0.025)(0.15)
2(0.2)(0.025) + 0.025(0.15) = 0.13068 m
2 m 1 m
150 mm
200 mm
25 mm
25 mm 25 mm
B
A
P
Absolute Maximum Bending Stress: The maximum tensile and compressive stressesoccur at the top and bottom-most fibers of the cross section. For the top fiber,
For the top fiber,
Ans.P = 2603.49 N = 2.60 kN (controls)
(sallow )c =M maxc
I 5(106) =
P (0.13068)
68.0457(10- 6)
P = 2885.79 N = 2.89 kN
4(106
) =
P(0.225 - 0.13068)
68.0457(10 - 6)(sallow )t =
M max y
I ;
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1149
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 103/135
*11–104. The member has a square cross section and issubjected to a resultant internal bending moment of
as shown. Determine the stress at eachcorner and sketch the stress distribution produced by M!
Set u = 45°.
M = 850 N # m
Ans.
Ans.
Ans.
Ans.
The negative sign indicates compressive stress.
sE = -601.04(0.125)
0.3255208(10 - 3)+
601.04(0.125)
0.3255208(10 - 3)= 0
sD = -601.04(0.125)
0.3255208(10 - 3)+
601.04( -0.125)
0.3255208(10 - 3)= -462kPa
sB = -601.04( -0.125)
0.3255208(10 - 3)+
601.04(0.125)
0.3255208(10 - 3)= 462kPa
sA = -601.04( -0.125)
0.3255208(10 - 3)+
601.04( -0.125)
0.3255208(10 - 3)= 0
s = -Mzy
Iz+
Myz
Iy
Iz = Iy =1
12(0.25)(0.25)3
= 0.3255208(10 - 3) m4
Mz = 850 sin 45° = 601.04 N # m
My = 850 cos 45° = 601.04 N # m
250 mm
125 mmB
A
z
y
E
M 850 NmC
125 mm
D
u
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1150
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 104/135
11–105. The member has a square cross section and issubjected to a resultant internal bending moment of
as shown. Determine the stress at eachcorner and sketch the stress distribution produced by M.Set u = 30°.
M = 850 N # m
Ans.
Ans.
Ans.
Ans.
The negative signs indicate compressive stress.
sE = -425(0.125)
0.3255208(10 - 3)+
736.12(0.125)
0.3255208(10 - 3)= 119 kPa
sD = -425(0.125)
0.3255208(10 - 3)+
736.12(- 0.125)
0.3255208(10 - 3)= - 446 kPa
sB = -425(-0.125)
0.3255208(10 - 3)+
736.12(0.125)
0.3255208(10 - 3)= 446 kPa
sA = -425(- 0.125)
0.3255208(10 - 3)+
736.12(- 0.125)
0.3255208(10 - 3)= - 119 kPa
s = -Mzy
Iz+
Myz
Iy
Iz = Iy =1
12 (0.25)(0.25)3
= 0.3255208(10 - 3) m4
Mz = 850 sin 30° = 425 N # m
My = 850 cos 30° = 736.12 N # m
250 mm
125 mmB
A
z
y
E
M 850 NmC
125 mm
D
u
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1151
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 105/135
11–106. Consider the general case of a prismatic beam
subjected to bending-moment components as
shown, when the x, y, z axes pass through the centroid of the
cross section. If the material is linear-elastic, the normal
stress in the beam is a linear function of position such
that . Using the equilibrium conditions
, ,
determine the constants a, b, and c, and show that the
normal stress can be determined from the equation
where the moments and products of inertia are defined in
Appendix A.
(IyIz - Iyz2),(MyIz + MzIyz)z]>MyIyz)y +[ -(MzIy +s =
Mz = LA
-y s dAMy = LA
z s dA0 = LA
s dA
s = a + by + cz
My and Mz,
Equilibrium Condition:
(1)
(2)
(3)
Section Properties: The integrals are defined in Appendix A. Note that
. Thus,
From Eq. (1)
From Eq. (2)
From Eq. (3)
Solving for a,b,c:
Thus, (Q.E.D.)sx = - ¢Mz Iy + My Iyz
Iy Iz - Iyz2 ≤y + ¢My Iy + MzIyz
Iy Iz - Iyz2 ≤z
b = - ¢MzIy + My Iyz
Iy Iz - Iyz2 ≤ c =
My Iz + Mz Iyz
Iy Iz - Iyz2
a = 0 (Since A Z 0)
Mz = -bIz - cIyz
My = bIyz + cIy
Aa = 0
LA y dA = LA z dA = 0
= -aLA ydA - bLA y2 dA - cLA yz dA
= LA -y(a + by + cz) dA
Mz = LA -y sx dA
= aLA z dA + bLA yz dA + cLA z2 dA
= LA z(a + by + cz) dA
My = LA z sx dA
0 = aLA dA + bLA y dA + cLA z dA
0 = LA
(a + by + cz) dA
0 = LA sx dA
sx = a + by + cz
y
y
z x
z
dA
M y
C
Mz
s
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1152
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 106/135
11–107. If the beam is subjected to the internal moment ofM = 2 kN m, determine the maximum bending stressdeveloped in the beam and the orientation of the neutral axis.
#
Internal Moment Components: The y and z components of M are positive since theyare directed towards the positive sense of their respective axes, Fig.a. Thus,
Section Properties: The location of the centroid of the cross-section is
The moments of inertia of the cross section about the principal centroidal y and z axes are
y =gy~ AgA =
0.025(0.05)(0.2) + 0.15(0.2)(0.05)
0.05(0.2) + 0.2(0.05) = 0.0875 m
Mz = 2 cos 60° = 1 kN # m
M y = 2 sin 60° = 1.732 kN # m
Bending Stress: By inspection, the maximum bending stress occurs at eithercorner A or B.
Ans.
Orientation of Neutral Axis: Here, .
Ans.
The orientation of the neutral axis is shown in Fig.b.
a = 79.8°
tan a =0.1135(10- 3)
35.4167(10 - 6) tan 60°
tan a =
I z
I y tan u
u = 60°
= 2.65 MPa (T)
sB = -1(103)(- 0.1625)
0.1135(10- 3)+
1.732(103)(0.025)
35.4167(10- 6)
= -5.66 MPa = 5.66 MPa (C) (Max.)
sA = -1(103)(0.0875)
0.1135(10 - 3)+
1.732(103)(-0.1)
35.4167(10- 6)
s = -Mzy
Iz
+
Myz
Iy
200 mm
50 mm
50 mm
y
z
A
B
x
y
100 mm
M
100 mm
60°
= 0.1135(10- 3) m4
+ 0.05(0.2)(0.15 - 0.0875)2+
1
12(0.05)(0.23)Iz =
1
12 (0.2)(0.053) + 0.2(0.05)(0.0875 - 0.025)2
Iy =1
12(0.05)(0.23) +
1
12 (0.2)(0.053) = 35.4167(10 - 6) m4
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1153
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 107/135
*11–108. If the wood used for the T-beam has an allowabletensile and compressive stress of and
respectively, determine the maximumallowable internal moment M that can be applied to thebeam.
(sallow)c = 6 MPa,(sallow)t = 4 MPa
Internal Moment Components: The y and z components of M are positive since theyare directed towards the positive sense of their respective axes, Fig.a.Thus,
Section Properties: The location of the centroid of the cross section is
The moments of inertia of the cross section about the principal centroidal y andz axes are
y = ©y~A©A
= 0.025(0.05)(0.2) + 0.15(0.2)(0.05)0.05(0.2) + 0.2(0.05)
= 0.0875 m
Mz = M cos 60° = 0.5M
My = M sin 60° = 0.8660M
= 0.1135(10- 3) m4
+ 0.05(0.2)(0.15 - 0.0875)2+
1
12(0.05)(0.23)Iz =
1
12(0.2)(0.053) + 0.2(0.05)(0.0875 - 0.025)2
Iy =1
12 (0.05)(0.23) +
1
12 (0.2)(0.053) = 35.417(10 - 6) m4
Bending Stress: By inspection, the maximum bending stress can occur at either
corner A or B. For corner A, which is in compression,
Ans.
For corner B which is in tension,
M = 3014.53 N # m = 3.01 kN # m
4(106) = - 0.5M(-
0.1625)0.1135(10- 3)
+ 0.8660M(0.025)35.417(10- 6)
sB = (sallow)t = -M z yB
I z+
M yzB
I y
M = 2119.71 N # m = 2.12 kN # m (controls)
- 6(106) = -0.5M(0.0875)
0.1135(10- 3)+
0.8660M(- 0.1)
35.417(10 - 6)
sA = (sallow)c = -M z y A
I z+
M yz A
I y
200 mm
50 mm
50 mm
y
z
A
B
x
y
100 mm
M
100 mm
60°
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1154
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 108/135
11–109. The box beam is subjected to the internal momentof M = 4 kN m, which is directed as shown. Determine themaximum bending stress developed in the beam and theorientation of the neutral axis.
#
Internal Moment Components: The y component of M is negative since it is directedtowards the negative sense of the y axis, whereas the z component of M which isdirected towards the positive sense of the z axis is positive, Fig.a. Thus,
Section Properties: The moments of inertia of the cross section about the principalcentroidal y and z axes are
Bending Stress: By inspection, the maximum bending stress occurs at corners A and D.
Ans.
Ans.
Orientation of Neutral Axis: Here, .
Ans.
The orientation of the neutral axis is shown in Fig.b.
a = -76.0°
tana =
0.2708(10- 3)
67.7083(10 - 6) tan (-
45°)
tan a =Iz
Iy tan u
u = -45°
= 4.70 MPa (T)
smax = sD = -2.828(103)(-0.15)
0.2708(10- 3)+
(-2.828)(103)(-0.075)
67.7083(10 - 6)
= -4.70 MPa = 4.70 MPa (C)
smax = s A = -2.828(103)(0.15)
0.2708(10- 3
)
+(- 2.828)(103)(0.075)
67.7083(10- 6
)
s = -Mzy
Iz+
Myz
Iy
Iz =1
12(0.15)(0.33) -
1
12(0.1)(0.23) = 0.2708(10- 3) m4
Iy =1
12(0.3)(0.153) -
1
12(0.2)(0.13) = 67.7083(10 - 6) m4
Mz = 4 cos 45° = 2.828 kN # m
My = - 4 sin 45° = -2.828 kN # m
z
y
x
150 mm
150 mm
25 mm50 mm
50 mm
50 mm
50 mm
45
25 mm
M
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1155
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 109/135
11–110. If the wood used for the box beam has anallowable bending stress of , determinethe maximum allowable internal moment M that can beapplied to the beam.
(sallow) = 6 MPa
Internal Moment Components: The y component of M is negative since it is directedtowards the negative sense of the y axis, whereas the z component of M, which isdirected towards the positive sense of the z axis, is positive, Fig.a. Thus,
Section Properties: The moments of inertia of the cross section about the principalcentroidal y and z axes are
Bending Stress: By inspection, the maximum bending stress occurs at corners A and D.Here,we will consider cornerD.
Ans.M = 5106.88 N # m = 5.11 kN # m
6(106) = -0.7071M(- 0.15)
0.2708(10 - 3)+
(- 0.7071M)(- 0.075)
67.708(10- 6)
sD = s allow = -M z yD
I z+
M yzD
I y
Iz =1
12(0.15)(0.33) -
1
12(0.1)(0.23) = 0.2708(10 - 3) m4
Iy =1
12(0.3)(0.153) -
1
12(0.2)(0.13) = 67.708(10- 6) m4
Mz = M cos 45° = 0.7071M
My = -M sin 45° = - 0.7071M
z
y
x
150 mm
150 mm
25 mm50 mm
50 mm
50 mm
50 mm
45
25 mm
M
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1156
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 110/135
Internal Moment Components: The y component of M is positive since it is directedtowards the positive sense of the y axis, whereas the z component of M, which isdirected towards the negative sense of the z axis, is negative, Fig.a. Thus,
Section Properties: The location of the centroid of the cross-section is given by
The moments of inertia of the cross section about the principal centroidal y and z
axes are
Bending Stress: By inspection, the maximum bending stress occurs at eithercorner A or B.
Ans.
Orientation of Neutral Axis: Here, .
Ans.
The orientation of the neutral axis is shown in Fig.b.
a = -66.5°
tan a =
5.2132 A10- 3 B
1.3078 A10- 3 B tan ( -30°)
tan a =
Iz
Iy tan u
u = -30°
= -131 MPa = 131 MPa (C)(Max.)
sB = -
c -1039.23 A103 B d(-0.3107)
5.2132 A10- 3 B +
600 A103 B( -0.15)
1.3078 A10- 3 B
= 126 MPa (T)
sA = - c-
1039.23 A10
3
B d(0.2893)
5.2132 A10- 3 B +
600 A103 B(0.15)
1.3078 A10- 3 B
s = -Mzy
Iz+
Myz
Iy
= 5.2132 A10 - 3 Bm4
- c 1
12(0.15) A0.153 B + 0.15(0.15)(0.375 - 0.2893)2 d
Iz =1
12 (0.3) A0.63 B + 0.3(0.6)(0.3 - 0.2893)2
Iy =1
12 (0.6) A0.33 B -
1
12 (0.15) A0.153 B = 1.3078 A10- 3 B m4
y =©yA
©A =
0.3(0.6)(0.3) - 0.375(0.15)(0.15)
0.6(0.3) - 0.15(0.15) = 0.2893 m
Mz = -1200 cos 30° = -1039.23 kN # m
My = 1200 sin 30° = 600 kN # m
11–111. If the beam is subjected to the internal moment of M = 1200 kN m, determine the maximum bending stressacting on the beam and the orientation of the neutral axis.
#
150 mm
150 mm
150 mm
150 mm
300 mm
150 mm
y
xz
M
30
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1157
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 111/135
Internal Moment Components: The y component of M is positive since it is directedtowards the positive sense of the y axis, whereas the z component of M, which isdirected towards the negative sense of the z axis, is negative, Fig.a. Thus,
Section Properties: The location of the centroid of the cross section is
The moments of inertia of the cross section about the principal centroidal y and z
axes are
Bending Stress: By inspection, the maximum bending stress can occur at eithercorner A or B. For corner A which is in tension,
Ans.
For corner B which is in compression,
M = 1376 597.12 N # m = 1377 kN # m
-150 A106 B = -(-0.8660M)(-0.3107)
5.2132 A10- 3 B +
0.5M(-0.15)
1.3078 A10 - 3 B
sB
= (sallow
)c
= -Mz yB
Iz+
My zB
Iy
M = 1185 906.82 N # m = 1186 kN # m (controls)
125 A106 B = -(-0.8660M)(0.2893)
5.2132 A10 - 3 B +
0.5M(0.15)
1.3078 A10- 3 B
sA = (sallow)t = -Mz yA
Iz+
My zA
Iy
= 5.2132 A10 - 3 Bm4
- c 1
12 (0.15) A0.153 B + 0.15(0.15)(0.375 - 0.2893)2 d
Iz =1
12 (0.3) A0.63 B + 0.3(0.6)(0.3 - 0.2893)2
Iy =1
12 (0.6) A0.33 B -
1
12 (0.15) A0.153 B = 1.3078 A10- 3 B m4
y =©yA
©A =
0.3(0.6)(0.3) - 0.375(0.15)(0.15)
0.6(0.3) - 0.15(0.15) = 0.2893 m
Mz = -M cos 30° = -0.8660M
My = M sin 30° = 0.5M
*11–112. If the beam is made from a material havingan allowable tensile and compressive stress of
and , respectively,determine the maximum allowable internal momentM thatcan be applied to the beam.
(sallow)c = 150 MPa(sallow)t = 125 MPa
150 mm
150 mm
150 mm
150 mm
300 mm
150 mm
y
xz
M
30
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1158
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 112/135
Ans.
Ans.
Ans.sA = Mc
I = =
When u = 0°
a = - 25.3°
tan a = tan (- 3°)tan a =
I z
I y tan u;
sA = -
998630(-
+
- -
=
s = -
M z y
I z+
Myz
Iy
Iy =
1
12
(150)(50
3
)=
Iz =
1
12
(50)(150
3
) =
;
My = - 00 sin 3° = -
Mz = 00 cos 3° = #998.63 N m10
52.33 N-m10
1.5625(10 ) mm
6
75) 25)52330(12)(6.187 MPa
14.(10 )
1000000(12)(75)5.36 MPa
46 414(10 ) mm
1.5625(10 )6
1.5625(10 )6
6
14.(10 )6
14 (10 )6
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
11–113. The board is used as a simply supported floor joist.If a bending moment of M = 1000 N · m is applied 3° fromthe z axis, determine the stress developed in the board at thecorner A. Compare this stress with that developed by the samemoment applied along the z axis (u = 0°). What is the angle a for the neutral axis when u = 3°? Comment: Normally, floorboards would be nailed to the top of the beams so that u ≈ 0°and the high stress due to misalignment would not occur.
M 1000 Nm
150 mm
y
x
z
50 mm
A
u 3
1159
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 113/135
Ans.
Ans.a = -63.1°
tan a = I z
I y tan u = tan ( -60°)
sB = + = -1 .
sD =-(-7.5) ( -
+(-
= - 2. 2
sA =-( -7.5) (75)
+
(150)= .3
s = -
Mzy
Iz+
Myz
Iy
Iz =1
12(300)( 3
+2
+1
12(50)( 3
+2
=
Iy =1
12 (50)( 3
+1
12 (200)( 3
=
y =
(25)(300)(50) + (150)(200)(50)
+=
Mz = - cos 60° = -
My = sin 60° = N # mm
N-mm
75 mm200(50)300(50)
50 )300 )
50 ) 300(50)(50 ) 200 ) 50(200)(75 )
21 4 MPa130.2
175) 25)9 MPa
2 7 MPa
1
15(10 )6 13(10 )6
67.5(10 )15(10 )6
6 41 10 ) mm14.58(
6 41 10 ) mm30.2(
(10 )6
(10 )6 61 10 )14.58(
610 )13(
130.2(10 )6 61 10 )14.58(
610 )13((10 )6
( 150)
61 10 )14.58(
610 )13(-(-7.5) (75)
130.2
(10 )6
(10 )6
130.2(10 )6
61 10 )14.58(
-
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
11–114. The T-beam is subjected to a bending moment ofM = 15 kN · m directed as shown. Determine the maximumbending stress in the beam and the orientation of the neutralaxis. The location y of the centroid,C , must be determined.
200 mm
50 mm
50 mm
– y
y
z
M 15 kNm
C
60
150 mm150 mm
1160
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 114/135
11–115. The beam has a rectangular cross section. If it issubjected to a bending moment of M = 3500 N m directedas shown, determine the maximum bending stress in thebeam and the orientation of the neutral axis.
#
Ans.
Ans.
Ans.a = -66.6°
tan a4
= I z
I y tan u =
3.375 (10- 4)
8.4375(10 - 5) tan ( -30°)
sD = 0.2084 MPa
sC = --3031.09(0.15)
0.3375(10- 3)+
1750( -0.075)
84.375(10- 6)= -0.2084 MPa
sB = --3031.09( -0.15)
0.3375(10 - 3)+
1750( -0.075)
84.375(10- 6)= -2.90 MPa (max)
sA = --3031.09(0.15)
0.3375(10- 3)+
1750(0.075)
84.375(10 - 6)= 2.90 MPa (max)
s = - Mzy
Iz+Myz
Iy
Iz =1
12(0.15)(0.33) = 0.3375(10- 3) m4
Iy =1
12(0.3)(0.153) = 84.375(10 - 6) m4
Mz = -3500 cos 30° = -3031.09 N # m
My = 3500 sin 30° = 1750 N # m
150 mm
150 mmz
y
x
M 3500 Nm
30
150 mm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1161
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 115/135
*11–116. For the section, 114(10- 6) m ,4Iz¿ =Iy¿ = 31.7(10 ) m ,- 6 4
Ans.=-2461.26( -0.14835)
117(10 - 6)+
438.42( -0.034519)
29.0(10- 6)= 2.60 MPa (T)
sA =-Mzy
Iz+
Myz
I y
z = 0.14 sin 10.1° - 0.06 cos 10.1° = -0.034519 m
y = -0.06 sin 10.1° - 0.14 cos 10.1° = -0.14835 m
Mz = 2500 cos 10.1° = 2461.26 N # m
My = 2500 sin 10.1° = 438.42 N # m
Iy = 29.0(10- 6) m4Iz = 117(10- 6) m4
60 mm
60 mm
60 mm 60 mm
140 mm
80 mm
z¿
y¿
10.10
M 2500 Nm C
A
z
y
Using the techniques outlined inAppendix A, the member’s cross-sectional area has principalmoments of inertia of and Iz =Iy = 29.0(10- 6) m4
Iy¿z¿ = 15.1(10- 6) m4.
computed about the principal axes of inertia y and z, respectively. If the section is subjected to a moment of
directed as shown, determine the stressproduced at point A, using Eq.11–17.M = 2500 N # m
117(10 - 6) m4,
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1162
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 116/135
11–117. Solve Prob. 11–116 using the equation developed inProb. 6–106.
60 mm
60 mm
60 mm 60 mm
140 mm
80 mm
z¿
y¿
10.10
M 2500 Nm C
A
z
y
Ans.=-32500(31.7)(10- 6) + 04(-0.14) + 30 + 2500(15.1)(10- 6)4( -0.06)
31.7(10- 6)(114)(10- 6) - 3(15.1)(10- 6)42 = 2.60 MPa (T)
sA =
-(Mz¿Iy¿ + My¿Iy¿z¿)y¿ + (My¿Iz¿ + Mz¿Iy¿z¿)z¿
Iy¿Iz¿ - Iy¿z¿2
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1163
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 117/135
11–118. If the applied distributed loading ofbe assumed to pass through the centroid of the beam’s crosssectional area, determine the absolute maximum bendingstress in the joist and the orientation of the neutral axis.Thebeam can be considered simply supported at A and B.
w = 4 kN>m can
Internal Moment Components: The uniform distributed load w can be resolved intoits y and z components as shown in Fig. a.
w y and wz produce internal moments in the beam about the z and y axes,
respectively. For the simply supported beam subjected to the uniform distributed
load, the maximum moment in the beam is Thus,
As shown in Fig.b, and are positive since they are directed towardsthe positive sense of their respective axes.
Section Properties: The moment of inertia of the cross section about the principalcentroidal y and z axes are
Bending Stress: By inspection,the maximum bending stress occurs at points A and B.
Ans.
Ans.
Orientation of Neutral Axis: Here,
Ans.
The orientation of the neutral axis is shown in Fig.c.
a = 72.5°
tan a =29.8192(10 - 6)
2.5142(10- 6) tan 15°
tan a = I z
I y tan u
u = tan - 1 c (M y)max
(M z)max
d = tan - 1 a 4.659
17.387b = 15°.
= -150.96 MPa = 151 MPa (C)
smax = s B = -
17.387(103)(0.1)
29.8192 (10- 6)+
4.659(103)(- 0.05)
2.5142(10- 6)
= 150.96 MPa = 151 MPa (T)
smax = s A = -
17.387(103)(- 0.1)
29.8192 (10- 6)+
4.659(103)(0.05)
2.5142(10 - 6)
s = -(Mz)max y
I z+
(M y )maxz
I y
Iz =1
12 (0.1)(0.23) -
1
12 (0.09)(0.173) = 29.8192(10 - 6) m4
Iy = 2 c 1
12 (0.015)(0.13) d +
1
12 (0.17)(0.013) = 2.5142(10 - 6) m4
(My)max(Mz)max
(My)max =
wzL2
8 =
1.035(62)
8 = 4.659 kN # m
(Mz)max =
w yL2
8 =
3.864(62)
8 = 17.387 kN # m
M max = wL2
8.
wz = 4 sin 15° = 1.035 kN>m
wy = 4 cos 15° = 3.864 kN>m
A
B
6 m
(6 m)
15
15
15
15
w
w
100 mm
100 mm
100 mm
15 mm
15 mm10 mm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1164
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 118/135
Internal Moment Components: The uniform distributed load w can be resolved intoits y and z components as shown in Fig. a.
w y and wz produce internal moments in the beam about the z and y axes,respectively. For the simply supported beam subjected to the a uniform distributed
load, the maximum moment in the beam is . Thus,
As shown in Fig. b, and are positive since they are directedtowards the positive sense of their respective axes.
Section Properties: The moment of inertia of the cross section about the principalcentroidal y and z axes are
Bending Stress: By inspection, the maximum bending stress occurs at points A and B.We will consider point A.
Ans.w = 4372.11 N>m = 4.37 kN>m
165(106) = -4.3467w(-0.1)
29.8192(10 - 6)+
1.1647w(0.05)
2.5142(10 - 6)
s A = sallow = -
(M z)max y A
I z +
(M y)maxz A
I y
Iz =1
12 (0.1)(0.23) -
1
12 (0.09)(0.173) = 29.8192(10- 6) m4
Iy = 2 c 1
12 (0.015)(0.13) d +
1
12 (0.17)(0.013) = 2.5142(10- 6) m4
(My)max(Mz)max
(My)max = wzL
2
8 =
0.2588w(62)
8 = 1.1647w
(Mz)max = w yL
2
8 =
0.9659w(62)
8 = 4.3476w
Mmax = wL2
8
wz = w sin 15° = 0.2588w
w y = w cos 15° = 0.9659w
11–119. Determine the maximum allowable intensityw of the uniform distributed load that can be applied to thebeam.Assume w passes through the centroid of the beam’scross sectional area and the beam is simply supported at Aand B. The beam is made of material having an allowablebending stress of .sallow = 165 MPa
A
B
6 m
(6 m)
15
15
15
15
w
w
100 mm
100 mm
100 mm
15 mm
15 mm10 mm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1165
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 119/135
From Fig. 6-43,
Ans.r = 0.2(40) = 8.0 mm
r
h = 0.2
w
h = 60
40 = 1.5
K = 1.46
120(106) = K J (153)(0.02)
112(0.007)(0.04)3 K
smax = K Mc
I
60 mm40 mm 7 mm
M M
r
r
11–120. The bar is subjected to a moment ofDetermine the smallest radius r of the fillets so that anallowable bending stress of is not exceeded.sallow = 120 MPa
M = 153 N m.#
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1166
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 120/135
From Fig. 6–43,
Ans.smax = K Mc
I = 1.57
J
17.5(0.02)112(0.007)(0.04)3
K = 14.7 MPa
K = 1.57
r
h =
6
40 = 0.15
w
h =
60
40 = 1.5;
11–121. The bar is subjected to a moment of If r = 6 mm determine the maximum bending stress in thematerial.
M = 17.5 N m.# 60 mm40 mm 7 mm
M M
r
r
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1167
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 121/135
80 mm
20 mm7 mm
M M
r
r
Allowable Bending Stress:
Stress Concentration Factor: From the graph in the text
with and , then .
Ans.r = 5.00 mm
r
20 = 0.25
r
h = 0.25K = 1.45
w
h =
80
20 = 4
K = 1.45
124 A106 B = KB 40(0.01)1
12 (0.007)(0.023)R
sallow = KMc
I
*11–122. The bar is subjected to a moment ofDetermine the smallest radius r of the fillets so
that an allowable bending stress of is notexceeded.
sallow = 124 MPa40 N # m.
M = 80 mm
20 mm7 mm
M M
r
r
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1168
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 122/135
Stress Concentration Factor: From the graph in the text with
and , then .
Maximum Bending Stress:
Ans.= 54.4 MPa
= 1.45
B17.5(0.01)
112 (0.007)(0.023) R
smax = KMc
I
K = 1.45r
h =
5
20 = 0.25
w
h =
80
20 = 4
11–123. The bar is subjected to a moment of If determine the maximum bending stress in thematerial.r = 5 mm,
M = 17.5 N m.# 80 mm
20 mm7 mm
M M
r
r
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1169
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 123/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
From Fig. 6-44.
Ans. 122 lb
s
; 36 1.92 20(0.625)
112 (0.5)(1.25)3
1.92
0.25
0.125 2;
0.125
1.25 0.1
1.75 1.25
2 0.25
11–1 24. The simply supported notched bar is subjected totwo forces P. Determine the largest magnitude ofP that canbe applied without causing the material to yield.The materialis A-36 steel. Each notch has a radius of 0.125 in.
20 in. 20 in.
1.75 in.
0.5 in.
P P
1.25 in.
20 in. 20 in.500 mm 500 mm
42 mm
12 mm
P P
30 mm
500 mm 500 mm
M = 0.5 P
0.5 m
500mm
500mm
1000 mm
r = 3 mm.
b 42 – 30
2 6 mm
6
3 2;
3
30 0.1
250 = 1.92 0.5P (106)(15)112(12)(30)3
P = 0.469 kN
1170
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 124/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
From Fig. 6-44,
Ans.smax
1.92 2000(0.625)
112 (0.5)(1.25)3
29.5 ksi
1.92
0.25
0.125 2;
0.125
1.25 0.1
1.75 1.25
2 0.25
11–1 5 . The simply supported notched bar is subjected tothe two loads, each having a magnitude ofDetermine the maximum bending stress developed in thebar, and sketch the bending-stress distribution acting overthe cross section at the center of the bar. Each notch has aradius of 0.125 in.
100 lb.
20 in. 20 in.
1.75 in.
0.5 in.
P P
1.25 in.
20 in. 20 in.500 mm 500 mm
42 mm
12 mm
P P
30 mm
500 mm 500 mm
500 N
250 N · m
500 mm
500 N500 N
500mm
500mm
1000
mm
500 N500 N 500 N
the two loads, each having a magnitude of P = 500 N.
radiius of r = 3 mm.
6
3 2;
3
30 0.1
1.92 250(103)(15)1
12(12)(30)3 266.7 MPa
b 42 – 30
2 6 mm
2
1171
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 125/135
From Fig. 6-43,
Ans.L = 0.95 m = 950 mm
19.6875(106) =
175(0.2 + L
2)(0.03)112 (0.01)(0.063)
(sB)max = (sA)max =MB
c
I
(sA)max = KMAc
I = 1.5 c
(35)(0.02)1
12(0.01)(0.043) d = 19.6875 MPa
K = 1.5
w
h =
60
40 = 1.5
r
h =
7
40 = 0.175
*11–126. Determine the length L of the center portion of the bar so that the maximum bending stress at A,B,and C isthe same.The bar has a thickness of 10 mm.
200 mm 200 mm
7 mm40 mm60 mm
350 N
B A C
7 mm
L
2
L
2
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1172
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 126/135
Stress Concentration Factor:
For the smaller section with and , we have
obtained from the graph in the text.
For the larger section with and , we have
obtained from the graph in the text.
Allowable Bending Stress:
For the smaller section
Ans.
For the larger section
M = 254 N # m
200 A106 B = 1.77B M(0.015)1
12 (0.015)(0.033)R
smax = sallow = K Mc
I ;
M = 41.7 N # m (Controls ! )
200 A106 B = 1.2B M(0.005)1
12 (0.015)(0.013)R
smax = sallow = K Mc
I;
K = 1.77r
h =
3
30 = 0.1
w
h =
45
30 = 1.5
K = 1.2r
h =
6
10 = 0.6
w
h =
30
10 = 3
11–127. The stepped bar has a thickness of 15 mm.Determine the maximum moment that can be applied to itsends if it is made of a material having an allowable bendingstress of sallow = 200 MPa.
M
10 mm30 mm
45 mm
3 mm6 mm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1173
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 127/135
a
M = 0.0190 wL2
+ ©M = 0; M +1
2 w
L (0.385L)2a 1
3b(0.385L) -
2wL
27 (0.385L) = 0
x = A 4
27 L = 0.385 L
+ c ©Fy = 0; 2wL
27 -
1
2 w
Lx2 = 0
11–128. The compound beam consists of two segments thatare pinned together at B. Draw the shear and momentdiagrams if it supports the distributed loading shown.
2/ 3 L
A C
B
1/ 3 L
w
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1174
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 128/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
Problem 11-129
The beam is constructed from four pieces of wood, glued together as shown. If the internal bendingmoment is M = 120 kN·m, determine the maximum bending stress in the beam. Sketch a
three-dimensional view of the stress distribution acting over the cross section.
Given: bo 300 mm:= do 300 mm:=
bi 250 mm:= di 250 mm:=
M 120kN m⋅:=
Solution:
Section Property :
I1
12 bo do
3⋅ bi di
3⋅−⋅:=
Maximum Bending Stress: σM c⋅
I=
cmax 0.5do:=
σmax
M cmax⋅
I:=
σmax 51.51 MPa= Ans.
1175
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 129/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
Problem 11-13
For the section, I y =31.7(10-6) m4, I z = 114(10-6) m4, I yz = 15.1(10-6) m4
in Appendix A, the member's cross-sectional area has principal moments of inertia of I y' = 29(10-6
z' -6 4
is subjected to a moment of M
(a) using Eq. 6-11 and (b) using the equation developed in Prob. 6-111.
Given: M 2000N m⋅:= 10.10deg:=
b1 80mm:= b2 140mm:=
h1 60mm:= h2 60mm:=
Iy 31.7 106− m
4:= Iz 114 10
6− m4
:=
Iyz 15.1 106− m
4:=
Iy' 29.0 106− m
4:= Iz' 117 10
6− m4
:=
Solution: ' −:=
Coordinates of Point A : yA b2:= zA h2:=
y'A
z'A
⎛
⎝
⎞
⎠
cos '
sin '
sin ' −
cos '
⎛
⎝
⎞
⎠
yA
zA
⎛
⎝
⎞
⎠⋅:=
y'A
z'A
⎛
⎝
⎞
⎠
148.35
34.52
⎛
⎝
⎞
⎠mm=
a) Using Eq. 6-11
Internal Moment Components :
My' M sin ⋅:= Mz' M cos ⋅:=
Bending Stress:
σA
Mz' y'A⋅
Iz'−
My' z'A⋅
Iy'
⎛
⎝
⎞
⎠:= σA 2.079− MPa= (C) Ans
b) Using the equation developed in Prob. 6-111
Internal Moment Components :
My 0:= Mz M:=
Bending Stress: Using formula developed in Prob. 6-111.
D Iy Iz⋅ Iyz2
−:=
σA1
DMz My
Iy
Iyz
Iyz
Iz
⎛
⎝
⎞
⎠⋅
yA−
zA
⎛
⎝
⎞
⎠⋅:=
σA 2.086 MPa= − (C) Ans.
and I = 117(10 ) m , computed about the principal axes of inertia y' and z'
. Using the techniques outline
) m
= 2 kN·m directed as shown, determine the stress produced at point A,
, respectively. If the section
0
1176
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 130/135
Maximum Bending Stress: The moment of inertia about y axis must be determinedfirst in order to use flexure formula
Thus,
Ans.
Maximum Bending Stress: Using integration
Ans.smax = 0.410 N>mm2= 0.410 MPa
125 A103 B =smax
5 (1.5238) A106 B
125 A103 B =smax
5 B -
3
2 y2(100 - y)
32 -
8
15 y(100 - y)
52 -
16
105 (100 - y)
72R 2 100 mm
0
M =
smax
5 L100 mm
0
y22 100 - y dy
dM = 2[y(s dA)] = 2by c asmax
100 by d(2z dy) r
smax = McI
= 125(0.1)30.4762(10-6)
= 0.410 MPa
= 30.4762 A106 B mm4= 30.4762 A10-6 B m4
= 20B -
3
2 y2
(100 - y)32 -
8
15 y (100 - y)
52 -
16
105(100 - y)
72R 2 100 mm
0
= 20L100 mm
0
y22 100 - y dy
= 2L
100 mm
0
y2 (2z) dy
I = LA y2 dA
*11–13 . A shaft is made of a polymer having a parabolicupper and lower cross section. If it resists an internal momentof , determine the maximum bending stressdeveloped in the material (a) using the flexure formula and(b) using integration. Sketch a three-dimensional view of thestress distribution acting over the cross-sectional area.Hint:Themoment of inertia is determined using Eq. A–3 of Appendix A.
M = 125 N # m y
z
x
M 125 N· m
50 mm
100 mm
50 mm
y 100 – z
2/ 25
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1
1177
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 131/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
Problem 11-132
The beam is made from three boards nailed together as shown. If the moment acting on the crosssection is M = 650 N·m, determine the resultant force the bending stress produces on the top board.
Given: bf 290 mm:= tf 15 mm:=
tw 20 mm:= dw 125 mm:=
M 650 N m⋅:=
Solution: D dw tf :=
y yi Ai⋅ ⋅
Ai ⋅=
yc
bf tf ⋅ 0.5⋅ tf 2 dw tw⋅ 0.5dw tf ⋅
bf tf ⋅ 2dw tw⋅:=
y
c
44.933mm=
If 1
12 bf ⋅ tf
3⋅ bf tf ⋅ yc 0.5tf −
2⋅:=
Iw1
122tw ⋅ dw
3⋅ 2tw dw⋅ yc 0.5dw tf −
2⋅:=
I If Iw:= I 17990374.89mm4
=
Bending Stress: σ Mc
I⋅=
At B: cB yc tf −:= σB M
cB
I⋅:= σB 1.0815MPa=
At A: cA yc:= σA McA
I⋅:= σA 1.6235MPa=
Resultant Force: For the top board.
F 0.5 σA σB bf tf ⋅ ⋅:= F 5.883 kN= Ans.
1178
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 132/135
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
Ans.
c
Ans. 2 20 166
0; 20 166 22 0
20 2
0;
20 2 0
11–13 . Draw the shear and moment diagrams for the
beam and determine the shear and moment in the beam as
functions of x, where 0 6 ft .
6 ft 4 ft
2 kip/ ft
50 kipft
8 kip
x
1. 8 m.
94 – 30 x – V = 0
V = 94 – 30 x
94 – 243.6 – 30 x x2 – M = 0
M = –15 x2 + 94 x – 243.6
243.6 kN · m 30 kN/m40 kN
75 kN · m
94 kN
1.8 m 1.2 m
M (kN · m)
V (kN)
40
–243.6
–48
–123
30 x
94 kN
1.8 m 1.2 m
30 kN/m
75 kNm
40 kN
x
3
1179
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 133/135
Case (a):
Case (b):
Case (a) provides higher strength since the resulting maximum stress is less for agiven M and a.
Case (a) Ans.
Ans.¢smax =8.4853 M
a3 -
6M
a3 = 2.49aM
a3 b
smax = Mc
I =
Ma 1
2 2 ab0.08333 a4
=8.4853 M
a3
I = 2 c1
36a2
2 2 ab a
1
2 2 ab
3
+1
2 a
2
2 2 ab a
1
2 2 ab c a
1
2 2 ab a
1
3bd2
d = 0.08333 a4
smax = Mc
I =
M (a>2)112(a)4
=6M
a3
*11–134. A wooden beam has a square cross section asshown. Determine which orientation of the beam providesthe greatest strength at resisting the moment M. What is thedifference in the resulting maximum stress in both cases?
a
a
a a
M M
(a) (b)
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1180
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 134/135
11–135. Draw the shear and moment diagrams for the shaftif it is subjected to the vertical loadings of the belt, gear, andflywheel. The bearings at A and B exert only verticalreactions on the shaft.
A B
200 mm
450 N
150 N
300 N
200 mm400 mm 300 mm
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8
1181
7/23/2019 Solutions Manual_Chapter 11_Final.pdf
http://slidepdf.com/reader/full/solutions-manualchapter-11finalpdf 135/135
Internal Moment Components:
Section Property:
Maximum Bending Stress: By Inspection, Maximum bending stress occurs at Aand B. Applying the flexure formula for biaxial bending at point A
Ans.
Ansu 45°
cos u - sin u = 0
ds
du=
6M
a3 (-sin u + cos u) = 0
=6M
a3 (cos u + sin u)
= --M cos u (a2)
112 a4
+-Msin u (-a
2)1
12 a4
s = -MzyIz
+ My zIy
Iy = Iz =1
12 a4
Mz = -M cos u
My = -M sin u
11–136. The strut has a square cross section a by a and issubjected to the bending moment M applied at an angleas shown. Determine the maximum bending stress in termsof a, M , and . What angle will give the largest bendingstress in the strut? Specify the orientation of the neutralaxis for this case.
uu
u
M
x
z
y
a
a
! #$%&'() *+,-%./() 0(,.1 2'/% #.$ 3.+ 45678 299 &/:1.' &$'$&;$+8 <1/' =%.$&/%9 /' >&(.$-.$+ ,)+$& %99 -(>?&/:1. 9%@' %' .1$? -,&&$).9?
$A/'.8 B( >(&./() (C .1/' =%.$&/%9 =%? D$ &$>&(+,-$+E /) %)? C(&= (& D? %)? =$%)'E @/.1(,. >$&=/''/() /) @&/./): C&(= .1$ >,D9/'1$&8