7–7.

Determine the internal normal force, shear force, andmoment at points C and D in the simply-supported beam.Point D is located just to the left of the 2500-lb force.

SOLUTION

With reference to Fig. a, we have

a

a

Using these results and referring to Fig. b, we have

Ans.

Ans.

a Ans.

Also, by referring to Fig. c, we have

Ans.

Ans.

a Ans.

The negative sign indicates that VD acts in the opposite sense to that shown on thefree-body diagram.

MD = 7875 lb # ft2625(3) - MD = 0+ ©MD = 0;

VD = -125 lbVD + 2625 - 2500 = 0+ c ©Fy = 0;

ND = 0©Fx = 0;:+

MC = 6375 lb # ftMC + 500(3)(1.5) - 2875(3) = 0+ ©MC = 0;

VC = 1375 lb2875 - 500(3) - VC = 0+ c ©Fy = 0;

NC = 0©Fx = 0;:+

Ax = 0©Fx = 0;:+Ay = 2875 lb2500(3) + 500(6)(9) - Ay(12) = 0+ ©MB = 0;

By = 2625 lbBy(12) - 500(6)(3) - 2500(9) = 0+ ©MA = 0;

BAC D

3 ft 3 ft 3 ft

500 lb/ft

2500 lb

3 ft

*7–24.

SOLUTIONFree body Diagram: The support reactions at A need not be computed.

Internal Forces: Applying equations of equilibrium to segment BC, we have

Ans.

Ans.

a

Ans.MC = -302 kN # m

-24.011.52 - 12.0142 - 40 sin 60°16.32 - MC = 0+ ©MC = 0;

VC = 70.6 kN

VC - 24.0 - 12.0 - 40 sin 60° = 0+ c ©Fy = 0;

-40 cos 60° NC = 0 NC = -20.0 kN:+ ©Fx = 0;

Determine the internal normal force, shear force, andbending moment at point C.

A3 m 3 m

0.3 m

CB

8 kN/m40 kN

3 m

60°

-

*7–28.

SOLUTION

a

Ans.

Ans.

a

Ans.

Ans.

Ans.

a

Ans.MF = - 1231 lb # ft = - 1.23 kip # ft

+ ©MF = 0; 307.8(4) + MF = 0

VF = -308 lb

+ c ©Fy = 0; - 307.8 - VF = 0

:+ ©Fx = 0; NF = 0

ME = - 490 lb # ft

+ ©ME = 0; -ME - 245.2(2) = 0

+ c ©Fy = 0; VE = 245 lb

NE = -250 lb

:+ ©Fx = 0; -NE - 250 = 0

By = 245.2 lb

+ c ©Fy = 0; By - 120 - 500 sin 60° + 307.8 = 0

Bx = 250 lb

:+ ©Fx = 0; Bx - 500 cos 60° = 0

Cy = 307.8 lb

+ ©MB = 0; -120(2) - 500 sin 60°(3) + Cy (5) = 0

Determine the normal force, shear force, and moment atsections passing through points E and F. Member BC ispinned at B and there is a smooth slot in it at C.The pin at Cis fixed to member CD.

A

2 ft 2 ft

500 lb

80 lb/ ft350 lb ft

60

3 ft 2 ft 4 ft 2 ft1 ft

DE B F

C

7–57.

Draw the shear and bending-moment diagrams for each of thetwo segments of the compound beam.

A

C D

150 lb/ft

B

10 ft 4 ft2 ft 2 ft

SOLUTIONSupport Reactions: From FBD (a),

a

From FBD (b),

a

Shear and Moment Functions: Member AB.

For [FBD (c)],

Ans.

a

Ans.

For [FBD (d)],

Ans.

a

Ans.

For member CBD, [FBD (e)],

Ans.

a Ans.

For [FBD (f)],

Ans.

Ans.M = 52450 - 306x6 lb # ft

306.2518 - x2 - M = 0+ ©M = 0;

V + 306.25 = 0 V = 306 lb+ c ©Fy = 0;

2 ft<x ◊ 8 ft

918.75x - M = 0 M = 5919x6 lb # ft+ ©M = 0;

918.75 - V = 0 V = 919 lb+ c ©Fy = 0;

0 ◊ x<2 ft

M = 5-75.0x2+ 2100x - 147006 lb # ft

-150114 - x2a14 - x

2b - M = 0+ ©M = 0;

V = 52100 - 150x6 lb

V - 150114 - x2 = 0+ c ©Fy = 0;

12 ft<x ◊ 14 ft

M = 5875x - 75.0x26 lb # ft

M + 150xax

2b - 875x = 0+ ©M = 0;

V = 5875 - 150x6 lb

875 - 150x - V = 0+ c ©Fy = 0;

0 ◊ x<12 ft

Dy + 918.75 - 1225 = 0 Dy = 306.25 lb+ c ©Fy = 0;

1225162 - Cy182 = 0 Cy = 918.75 lb+ ©MD = 0;

Ay + 1225 - 2100 = 0 Ay = 875 lb+ c ©Fy = 0;

By 1122 - 2100172 = 0 By = 1225 lb+ ©MA = 0;

7–61.

Draw the shear and moment diagrams for the beam.

SOLUTION

Ans.Support Reactions: From FBD (a),

a

Shear and Moment Functions: For [FBD (b)],

Ans.

The maximum moment occurs when then

a

Ans.

Thus,

Mmax = 3.0013.4642 -

3.4643

12= 6.93 kN # m

M = b3.00x -

x3

12r kN # m

M + ¢x2

4≤ ax

3b - 3.00x = 0+ ©M = 0;

0 = 3.00 -

x2

4x = 3.464 m

V = 0,

V = b3.00 -

x2

4r kN

3.00 -

x2

4- V = 0+ c ©Fy = 0;

0 … x … 6 m

9.00122 - Ay162 = 0 Ay = 3.00 kN+ ©MB = 0;

3 m

6 m

A

C

B

3 kN/m

7–87.

SOLUTIONSupport Reactions:

a

Ay -

w0L

2-

w0L

4= 0 Ay =

3w0L

4+ c ©Fy = 0;

MA =

7w0L2

24

MA -

w0L

2aL

4b -

w0L

4a2L

3b = 0+ ©MA = 0;

Draw the shear and moment diagrams for the beam. w0

L––2

L––2

*7–92.

SOLUTIONSupport Reactions: From FBD (a),

a

Shear and Moment Diagrams: The value of the moment at supports A and B can beevaluated using the method of sections [FBD (c)].

a+ ©M = 0; M + 15.0(2) + 15 = 0 M = -45.0 kip # ft

Ay = 40.0 kip

+ c ©Fy = 0; Ay + 40.0 - 15.0 - 50.0 - 15.0 = 0

By = 40.0 kip

+ ©MA = 0; By (10) + 15.0(2) + 15 - 50.0(5) - 15.0(12) - 15 = 0

Draw the shear and moment diagrams for the beam.

A

6 ft 10 ft 6 ft

5 kip/ ft

B

15 kip ft15 kip ft

7–105.

SOLUTION

At ,

At ,

At ,

Ans.w = 51.9 lb/ft

FH = 2705 lb

Tmax =

FH

cos umax= 3000

umax = tan- 1 (0.48) = 25.64°

dy

dx2max

= tan umax =

wFH

x 2x = 25 ft

y = 6 ft FH = 52.08 wx = 25 ft

y =

w2FH

x2

C1 = C2 = 0

y = 0x = 0

dy

dx= 0x = 0

y =

1FHL aLwdxbdx

Determine the maximum uniform loading measured inthat the cable can support if it is capable of sustaining

a maximum tension of 3000 lb before it will break.lb>ft,

w, 50 ft

6 ft

w

7–106.

The cable is subjected to a uniform loading of Determine the maximum and minimum tension in the cable.

w = 250 lb>ft.

SOLUTIONFrom Example 7–12:

Ans.

The minimum tension occurs at .

Ans.Tmin = FH = 13.0 kip

u = 0°

Tmax =

FH

cos umax=

13 021cos 25.64°

= 14.4 kip

umax = tan- 1aw0 L

2FHb = tan- 1 a 250 (50)

2(13 021)b = 25.64°

FH =

w0 L2

8 h=

250 (50)2

8 (6)= 13 021 lb

50 ft

6 ft

w

7–111.

If the slope of the cable at support A is zero, determine thedeflection curve y = f(x) of the cable and the maximumtension developed in the cable.

SOLUTION

Using Eq. 7–12,

Applying the boundary condition at x = 0 results in C1 = 0.

Applying the boundary condition y = 0 at x = 0, we have

Thus,

Applying the boundary condition y = 4.5 m at x = 12 m, we have

Substituting this result into Eqs. (1) and (2), we obtain

and

Ans.

The maximum tension occurs at point B where the cable makes the greatest anglewith the horizontal. Here,

Thus,

Ans.Tmax =

FH

cos umax=

51.876(103)

cos 30.50°= 60.207(103) N = 60.2 kN

umax = tan-1ady

dx`x = 12 m

b = tan-1 c0.5890 sina p24

(12)b d = 30.50°

= 4.5a1 - cos p

24xb m

y =

2304(103)

p 2(51.876)(103)

c1 - cos p

24x d

= 0.5890 sin p

24x

dy

dx=

96(103)

p(51.876)(103) sin p

24x

FH = 51.876(103) N

4.5 =

2304(103)

p2FHc1 - cos

p

24 (12) d

y =

2304(103)

p2FHc1 - cos

p

24x d

C2 =

2304(103)

p2FH

0 = -

24pc96(103)

pFH cos 0° d + C2

dy

dx= 0

y = - 24p

c96(103)

pFH cos

p

24x d + C1x + C2

y =

1FHL

24pc4(103) d sin

p

24x + C1

y =

1FHL

aL

4 cos p

24* dxbdx

y =

1FHL

aL

w(x)dxbdx

12 m

4.5 m

4 kN/m

y

x

B

A

––24w � 4 cos xp

*7–120.

The power transmission cable weighs . If theresultant horizontal force on tower is required to bezero, determine the sag h of cable .

SOLUTION

The origin of the x, y coordinate system is set at the lowest point of the cables. Here,Using Eq. 4 of Example 7–13,

Applying the boundary condition of cable AB, at

Solving by trial and error yields

Since the resultant horizontal force at B is required to be zero,Applying the boundary condition of cable BC

Ans. = 4.44 ft

h =

11266.6210

c cosh B10(-100)

11266.62R - 1 s

y = h at x = -100 ft to Eq. (1), we obtain (FH)BC = (FH)AB = 11266.62 lb.

(FH)AB = 11266.63 lb

10 =

(FH)AB

10 Bcosh ¢10(150)

(FH)AB≤ - 1R

x = 150 ft,y = 10 ft

y =

FH

10 B cosh ¢ 10

FHx≤ - 1R ft

y =

FH

w0 B cosh ¢ w0

FHx≤ - 1R

w0 = 10 lb>ft.

BCBD

10 lb>ftA B

hC

D

300 ft

10 ft

200 ft

7–121.

The power transmission cable weighs . If ,determine the resultant horizontal and vertical forces thecables exert on tower .

SOLUTION

The origin of the x, y coordinate system is set at the lowest point of the cables. Here,Using Eq. 4 of Example 7–13,

Applying the boundary condition of cable AB, at

Solving by trial and error yields

Applying the boundary condition of cable BC, at to Eq. (2),we have

Solving by trial and error yields

Thus, the resultant horizontal force at B is

Ans.

Using Eq. (1), and

Thus, the vertical force of cables AB and BC acting

on point B are

The resultant vertical force at B is therefore

Ans. = 2511.07 lb = 2.51 kip

(Fv)R = (Fv)AB + (Fv)BC = 1504.44 + 1006.64

(Fv)BC = (FH)BC tan (uB)BC = 5016.58(0.20066) = 1006.64 lb

(Fv)AB = (FH)AB tan (uB)AB = 11266.63(0.13353) = 1504.44 lb

sin h B10(-100)

5016.58R = 0.20066.

tan (uB)BC = tan (uB)AB = sin h B 10(150)

11266.63R = 0.13353

(FH)R = (FH)AB - (FH)BC = 11266.63 - 5016.58 = 6250 lb = 6.25 kip

(FH)BC = 5016.58 lb

10 =

(FH)BC

10 Bcosh¢10(100)

(FH)BC≤ - 1R

x = -100 fty = 10 ft

(FH)AB = 11266.63 lb

10 =

(FH)AB

10 Bcos h ¢10(150)

(FH)AB≤ - 1R

x = 150 ft,y = 10 ft

y =

FH

10 B cos h ¢ 10

FHx≤ - 1R ft

y =

FH

w0 B cos h ¢ w0

FHx≤ - 1R

w0 = 10 lb>ft.

BD

h = 10 ft10 lb>ftA B

hC

D

300 ft

10 ft

200 ft

7–134.

Determine the normal force, shear force, and moment atpoints B and C of the beam.

5 m5 m 3 m

2 kN/m1 kN/m

7.5 kN

40 kN m

6 kN

1 m

A BC

SOLUTIONFree body Diagram: The support reactions need not be computed for this case .

Internal Forces: Applying the equations of equilibrium to segment DC [FBD (a)],we have

Ans.

Ans.

a

Ans.

Applying the equations of equilibrium to segment DB [FBD (b)], we have

Ans.

Ans.

a

Ans.MB = -184.5 kN # m

- 4.00(7) - 6(9) - 40 = 0

+ ©MB = 0; -MB - 10.0(2.5) - 7.5(5)

VB = 27.5 kN

+ c ©Fy = 0; VB - 10.0 - 7.5 - 4.00 - 6 = 0

:+ ©Fx = 0; NB = 0

MC = -62.5 kN # m

+ ©MC = 0; -MC - 3.00(1.5) - 6(3) - 40 = 0

+ c ©Fy = 0; VC - 3.00 - 6 = 0 VC = 9.00 kN

:+ ©Fx = 0; NC = 0