Solution manual for mechanics of materials 10th edition hibbeler sample

731

Solution

sallow =p r

2 t; 12(106) =

300(103)(1.5)

2 t

t = 0.0188 m = 18.8 mm Ans.

81.

A spherical gas tank has an inner radius of r = 1.5 m. If it is subjected to an internal pressure of p = 300 kPa, determine its required thickness if the maximum normal stress is not to exceed 12 MPa.

Ans: t = 18.8 mm

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732

.

Solution

sallow =p r

2 t; 15(103) =

200 ri2(0.5)

ri = 75 in.

ro = 75 in. + 0.5 in. = 75.5 in. Ans.

82.

A pressurized spherical tank is made of 0.5-in.-thick steel. If it is subjected to an internal pressure of p = 200 psi, determine its outer radius if the maximum normal stress is not to exceed 15 ksi.

Ans:ro = 75.5 in.

733

83.

The thin-walled cylinder can be supported in one of two ways as shown. Determine the state of stress in the wall of the cylinder for both cases if the piston P causes the internal pressure to be 65 psi. The wall has a thickness of 0.25 in. and the inner diameter of the cylinder is 8 in.

SolutionCase (a):

s1 =pr

t ; s1 =

65(4)

0.25= 1.04 ksi Ans.

s2 = 0 Ans.

Case (b):

s1 =pr

t ; s1 =

65(4)

0.25= 1.04 ksi Ans.

s2 =pr

2t ; s2 =

65(4)

2(0.25)= 520 psi Ans.

P

(a) (b)

P

8 in. 8 in.

Ans:(a) s1 = 1.04 ksi, s2 = 0, (b) s1 = 1.04 ksi, s2 = 520 psi

734

.

*84.

The tank of the air compressor is subjected to an internal pressure of 90 psi. If the inner diameter of the tank is 22 in., and the wall thickness is 0.25 in., determine the stress components acting at point A. Draw a volume element of the material at this point, and show the results on the element. A

Solution

Hoop Stress for Cylindrical Vessels: Since rt=

110.25

= 44 7 10, then thin wall

analysis can be used. Applying Eq. 81

s1 =pr

t=

90(11)

0.25= 3960 psi = 3.96 ksi Ans.

Longitudinal Stress for Cylindrical Vessels: Applying Eq. 82

s2 =pr

2t=

90(11)

2(0.25)= 1980 psi = 1.98 ksi Ans.

Ans:s1 = 3.96 ksi,s2 = 1.98 ksi

735

85.

Air pressure in the cylinder is increased by exerting forces P = 2 kN on the two pistons, each having a radius of 45 mm. If the cylinder has a wall thickness of 2 mm, determine the state of stress in the wall of the cylinder.

47 mm

P

P

Solution

p =PA

=2(103)

p(0.0452)= 314 380.13 Pa

s1 =p r

t=

314 380.13(0.045)

0.002= 7.07 MPa Ans.

s2 = 0 Ans.

The pressure P is supported by the surface of the pistons in the longitudinal direction.

Ans:s1 = 7.07 MPa, s2 = 0

736

.

86.

Determine the maximum force P that can be exerted on each of the two pistons so that the circumferential stress in the cylinder does not exceed 3 MPa. Each piston has a radius of 45 mm and the cylinder has a wall thickness of 2 mm.

47 mm

P

P

Solution

s =p r

t ; 3(106) =

p(0.045)

0.002

p = 133.3 kPa

P = pA = 133.311032 (p)(0.045)2 = 848 N Ans.

Ans:P = 848 N

737

Solution

a) s1 =pr

t=

1.35(106)(0.75)

0.008= 126.56(106) = 127 MPa Ans.

b) 126.56 (106)(0.05)(0.008) = s1(2)(0.04)(0.008)

s1 = 79.1 MPa Ans.

c) From FBD(a)

+ cFy = 0; Fb - 79.1(106)[(0.008)(0.04)] = 0

Fb = 25.3 kN

(tavg)b =FbA

-25312.5p4(0.01)

2 = 322 MPa Ans.

87.

A boiler is constructed of 8-mm-thick steel plates that are fastened together at their ends using a butt joint consisting of two 8-mm cover plates and rivets having a diameter of 10 mm and spaced 50 mm apart as shown. If the steam pressure in the boiler is 1.35 MPa, determine (a) the circumferential stress in the boilers plate away from the seam, (b) the circumferential stress in the outer cover plate along the rivet line aa, and (c) the shear stress in the rivets.

a

8 mm

50 mm a

0.75 m

Ans:(a) s1 = 127 MPa,(b) s1 = 79.1 MPa,(c) (tavg)b = 322 MPa

738

.

SolutionNormal Stress: Since the pipe has two open ends,

slong = s2 = 0 Ans.

Since rt=

60.25

= 24 7 10, thin-wall analysis can be used.

sh = s1 =pr

t=

250(6)

0.25= 6000 psi = 6 ksi Ans.

*88.

The steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. If the valve A is opened and the flowing water has a pressure of 250 psi as it passes point B, determine the longitudinal and hoop stress developed in the wall of the pipe at point B.

AB

Ans:slong = 0,sh = 6 ksi

739

Solution

Normal Stress: Since rt=

60.25

= 24 7 10, thin-wall analysis can be used.

shoop = s1 =pr

t=

300(6)

0.25= 7200 psi = 7.20 ksi Ans.

slong = s2 =pr

2t=

300(6)

2(0.25)= 3600 psi = 3.60 ksi Ans.

The state of stress on an element in the pipe wall is shown in Fig. a.

89.

The steel water pipe has an inner diameter of 12 in. and a wall thickness of 0.25 in. If the valve A is closed and the water pressure is 300 psi, determine the longitudinal and hoop stress developed in the wall of the pipe at point B. Draw the state of stress on a volume element located on the wall.

AB

Ans:shoop = 7.20 ksi, slong = 3.60 ksi

740

.

Solution

s1 =400

2(1>8)(1) = 1600 psi Ans.

s1 =pr

t ; 1600 =

p(8)

(1>8) p = 25 psi Ans.

P1 =s1

E=

1600

29(106)= 55.1724(10- 6)

d = P1L = 55.1724(10- 6)(p)a8 + 116 b = 0.00140 in. Ans.

810.

The A-36-steel band is 2 in. wide and is secured around the smooth rigid cylinder. If the bolts are tightened so that the tension in them is 400 lb, determine the normal stress in the band, the pressure exerted on the cylinder, and the distance half the band stretches.

8 in.

in.18

Ans:s1 = 1.60 ksi, p = 25 psi, d = 0.00140 in.

741

SolutionRequire

dF = dT ; dF =PLAE

=sLE

, dT = aTL

s2(20)(12)

29(106)= (6.60)(10- 6)(60)(20)(12)

s2 = 11.5 ksi Ans.

s1 =pr

t=

600(10)

0.25= 24 ksi Ans.

811.

The gas pipe line is supported every 20 ft by concrete piers and also lays on the ground. If there are rigid retainers at the piers that hold the pipe fixed, determine the longitudinal and hoop stress in the pipe if the temperature rises 60 F from the temperature at which it was installed. The gas within the pipe is at a pressure of 600 lb> in2. The pipe has an inner diameter of 20 in. and thickness of 0.25 in. The material is A-36 steel.

20 ft

Ans:s2 = 11.5 ksi,s1 = 24 ksi

742

.

Solution+cFy = 0; p(0.225)2450(103) - tavg (2p)(0.225)(0.01) = 0;

tavg = 5.06 MPa Ans.

s1 =p r

t=

450(103)(0.225)

0.02= 5.06 MPa Ans.

s2 =p r

2 t=

450(103)(0.225)

2(0.02) = 2.53 MPa Ans.

*812.

A pressure-vessel head is fabricated by welding the circular plate to the end of the vessel as shown. If the vessel sustains an internal pressure of 450 kPa, determine the average shear stress in the weld and the state of stress in the wall of the vessel.

450 mm10 mm

20 mm

Ans:tavg = 5.06 MPa,s1 = 5.06 MPa,s2 = 2.53 MPa

743

SolutiondT = a TL

p(24) - p(23.99) = 6.60(10- 6)(T1 - 65)(p)(23.99)

T1 = 128.16F = 128F Ans.

Cool down:

dF = dT

FLAE

= a TL

F(p)(24)

(1)(0.25)(29)(106)= 6.60(10- 6)(128.16 - 65)(p)(24)

F = 3022.21 lb

s1 =FA

; s1 =3022.21

(1)(0.25)= 12 088 psi = 12.1 ksi Ans.

s1 =pr

t ; 12 088 =

p(12)

(0.25)

P = 252 psi Ans.

813.

An A-36-steel hoop has an inner diameter of 23.99 in., thickness of 0.25 in., and width of 1 in. If it and the 24-in.-diameter rigid cylinder have a temperature of 65 F, determine the temperature to which the hoop should be heated in order for it to just slip over the cylinder. What is the pressure the hoop exerts on the cylinder, and the tensile stress in the ring when it cools back down to 65 F?

24 in.

Ans:T1 = 128F, s1 = 12.1 ksi, p = 252 psi

744

.

SolutionEquilibrium for the Ring: From the FBD

+S Fx = 0; 2P - 2pri w = 0 P = pri w

Hoop Stress and Strain for the Ring:

s1 =PA

=pri w

(ro - ri)w=

priro - ri

Using Hookes Law

P1 =s1

E=

priE(ro - ri)

(1)

However, P1 =2p(ri)1 - 2pri

2pri=

(ri)1 - riri

=driri

.

Then, from Eq. (1)

driri

=pri

E(ro - ri)

dri =pr i

2

E(ro - ri) Ans.

814.

The ring, having the dimensions shown, is placed over a flexible membrane which is pumped up with a pressure p. Determine the change in the inner radius of the ring after this pressure is applied. The modulus of elasticity for the ring is E.

p

ro

w

ri

Ans:

dri =pr i

2

E(ro - ri)

745

SolutionEquilibrium for the Ring: From the FBD

+S Fx = 0;2P - 2pri w = 0 P = pri w

Hoop Stress and Strain for the Ring:

s1 =PA

=pri w

(ro - ri)w=

priro - ri

Using Hookes Law

P1 =s1

E=

priE(ro - ri)

(1)

However, P1 =2p(ri)1 - 2pri

2pri=

(ri)1 - riri

=driri

.

Then, from Eq. (1)

driri

=pri

E(ro - ri)

dri =pr i

2

E(ro - ri)

Compatibility: The pressure between the rings requires

dr2 + dr3 = r2 - r3 (2)

From the result obtained above

dr2 =pr 2

2

E(r2 - r1) dr3 =

pr 32

E(r4 - r3)

Substitute into Eq. (2)

pr 2

2

E(r2 - r1)+

pr 32

E(r4 - r3)= r2 - r3

p =E(r2 - r3)

r 22r2 - r1

+r 3

2

r4 - r3

Ans.

815.

The inner ring A has an inner radius r1 and outer radius r2. The outer ring B has an inner radius r3 and an outer radius r4, and r2 7 r3. If the outer ring is heated and then fitted over the inner ring, determine the pressure between the two rings when ring B reaches the temperature of the inner ring. The material has a modulus of elasticity of E and a coefficient of thermal expansion of a.

r1

r2

r3

A B

r4

Ans:

p =E(r2 - r3)

r 22r2 - r1

+r 3

2

r4 - r3

746

.

SolutionNormal Pressure: Vertical force equilibrium for FBD(a).

+ cFy = 0; 103p(242)4 - N = 0 N = 5760p lbThe Friction Force: Applying friction formula

Ff = ms N = 0.5(5760p) = 2880p lb

a) The Required Torque: In order to initiate rotation of the two hemispheres relative to each other, the torque must overcome the moment produced by the friction force about the center of the sphere.

T = Ff r = 2880p(2 + 0.125>12) = 18190 lb # ft = 18.2 kip # ft Ans.b) The Required Vertical Force: In order to just pull the two hemispheres apart, thevertical force P must overcome the normal force.

P = N = 5760p = 18096 lb = 18.1 kip Ans.

c) The Required Horizontal Force: In order to just cause the two hemispheres toslide relative to each other, the horizontal force F must overcome the friction force.

F = Ff = 2880p = 9048 lb = 9.05 kip Ans.

*816.

Two hemispheres having an inner radius of 2 ft and wall thickness of 0.25 in. are fitted together, and the inside pressure is reduced to -10 psi. If the coefficient of static friction is ms = 0.5 between the hemispheres, determine (a) the torque T needed to initiate the rotation of the top hemisphere relative to the bottom one, (b) the vertical force needed to pull the top hemisphere off the bottom one, and (c) the horizontal force needed to slide the top hemisphere off the bottom one.

Ans: (a) T = 18.2 kip # ft,(b) P = 18.1 kip,(c) F = 9.05 kip

747

.

SolutionNormal Stress in the Wall and Filament Before the Internal Pressure is Applied: The entire length L of wall is subjected to pretension filament force T. Hence, from equilibrium, the normal stress in the wall at this state is

2T - (s)w (2Lt) = 0 (s)w =TLt

and for the filament the normal stress is

(s)fil =T

wt

Normal Stress in the Wall and Filament After the Internal Pressure is Applied: In order to use s1 = pr>t, developed for a vessel of uniform thickness, we redistribute the filaments cross-section as if it were thinner and wider, to cover the vessel with no gaps. The modified filament has width L and thickness t>w>L, still with cross-sectional area wt> subjected to tension T. Then the stress in the filament becomes

sfil = s + (s)fil =pr

(t + tw>L) +T

wtAns.

And for the wall,

sw = s - (s)w =pr

(t + tw>L) -TLt

Ans.

Check: 2wtsfil + 2Ltsw = 2rLp OK

817.

In order to increase the strength of the pressure vessel, filament winding of the same material is wrapped around the circumference of the vessel as shown. If the pretension in the filament is T and the vessel is subjected to an internal pressure p, determine the hoop stresses in the filament and in the wall of the vessel. Use the free-body diagram shown, and assume the filament winding has a thickness t and width w for a corresponding length L of thevessel.

T

p

w

t

L

t

T

s1

s1

Ans:

sfil =pr

t + tw>L +T

wt,

sw =pr

t + tw>L -TLt

748

.

818.

Determine the shortest distance d to the edge of the plate at which the force P can be applied so that it produces no compressive stresses in the plate at section aa. The plate has a thickness of 10 mm and P acts along the centerline of this thickness.

SolutionsA = 0 = sa - sb

0 =PA

-M c

I

0 =P

(0.2)(0.01)-

P(0.1 - d)(0.1)112 (0.01)(0.2

3)

P(-1000 + 15000 d) = 0

d = 0.0667 m = 66.7 mm Ans.

a

500 mm

P

a

300 mm

200 mm

d

Ans: d = 66.7 mm

749

.

819.

Determine the maximum distance d to the edge of the plate at which the force P can be applied so that it produces no compressive stresses on the plate at section aa. The plate has a thickness of 20 mm and P acts along the centerline of this thickness.

SolutionInternal Loadings: Consider the equilibrium of the left segment of the plate sectioned through section aa, Fig. a,

S+ Fx = 0; N - P = 0 N = P

a+Mo = 0; P(d - 0.1) - M = 0 M = P(d - 0.1)

Section Properties: For the rectangular cross section,

A = 0.2(0.02) = 0.004 m2

I =112

(0.02)(0.23) = 13.3333(10-6) m4

Normal Stress: It is required that sA = 0. For the combined loadings,

sA =NA

-McI

0 =P

0.004-

P(d - 0.1)(0.1)13.3333(10-6)

d = 0.1333 m = 133 mm Ans.

a

a

P

d

200 mm

Ans:d = 133 mm

750

.

*820.

The plate has a thickness of 20 mm and the force P = 3 kN acts along the centerline of this thickness such that d = 150 mm. Plot the distribution of normal stress acting along section aa.

SolutionInternal Loadings: Consider the equilibrium of the left segment of the plate sectioned through section aa, Fig. a,

+S Fx = 0; N - 3 = 0 N = 3.00 kN

a+Mo = 0; 3(0.05) - M = 0 M = 0.150 kN # mSection Properties: For the rectangular cross section,

A = 0.2(0.02) = 0.004 m2

I =112

(0.02)(0.23) = 13.3333(10-6) m4

Normal Stress: For the combined loadings, the normal stress at points A and B can be determined from

s =NA

{ McI

=3.00(103)

0.004{

0.150(103)(0.1)13.3333(10-6)

sA = 750(103) - 1.125(106) = -0.375(106) Pa = 0.375 MPa (C) Ans.

sB = 750(103) + 1.125(106) = 1.875(106) Pa = 1.875 MPa (T) Ans.

Using similar triangles, the location of the neutral axis can be determined

y

1.875=

0.2 - y0.375

; y = 0.1667 m

a

a

P

d

200 mm

Ans:sA = 0.375 MPa (C),sB = 1.875 MPa (T)

751

.

821.

If the load has a weight of 600 lb, determine the maximum normal stress on the cross section of the supporting member at section aa. Also, plot the normal-stress distribution over the cross section.

SolutionInternal Loadings: Consider the equilibrium of the free-body diagram of the bottom cut segment shown in Fig. a.

a+ cFy = 0; N - 600 = 0 N = 600 lb

a+MC = 0; 600(1.5) - M = 0 M = 900 lb # ftSection Properties: The cross-sectional area and the moment of inertia about the centroidal axis of the member are

A = p(12) = p in2 I =p

4 (14) =

p

4 in4

Normal Stress: The normal stress is the combination of axial and bending stress. Thus,

s =NA

{ McI

By observation, the maximum normal stress occurs at point B, Fig. b. Thus,

smax = sB =600p

+900(12)(1)

p>4 = 13.9 ksi (T) Ans.

For Point A,

sA =600p

+ - 900(12)(1)

p>4 = -13.6 ksi = 13.6 ksi (C) Ans.

Using these results, the normal stress distribution over the cross section is shown in Fig. b. The location of the neutral axis can be determined from

p

13.9=

2 - x13.6

; x = 1.01 in.

Ans:smax = sB = 13.9 ksi (T), sA = 13.6 ksi (C)

1.5 ft1 in.a a

Section a a

752

.

822.

The steel bracket is used to connect the ends of twocables. If the allowable normal stress for the steel is sallow = 30 ksi, determine the largest tensile force P that can be applied to the cables. Assume the bracket is a rod having a diameter of 1.5 in.

SolutionInternal Loading: Consider the equilibrium of the left segment of the bracket sectioned through an arbitrary cross section, Fig. a.

+S Fx = 0; N - P = 0 N = P

a+Mo = 0; M - P(4.75) = 0 M = 4.75P

Section Properties: For the circular cross section,

A = pc2 = p(0.752) = 0.5625p in2

I =p

4c4 =

p

4(0.754) = 0.24850 in4

Allowable Normal Stress: The maximum normal stress occurs at the bottom of the circular cross section. For the combined loadings,

smax = sallow =NA

+McI

30(103) = P0.5625p

+4.75P(0.75)

0.24850

P = 2.01320(103) lb

= 2.01 kip Ans.

4 in.

P P

Ans:Pmax = 2.01 kip

753

.

823.

The steel bracket is used to connect the ends of two cables. If the applied force P = 1.50 kip, determine the maximum normal stress in the bracket. Assume the bracket is a rod having a diameter of 1.5 in.

SolutionInternal Loading: Consider the equilibrium of the left segment of the bracket sectioned through an arbitrary cross section, Fig. a.

+S Fx = 0; N - 1.50 = 0 N = 1.50 kip

a+Mo = 0; M - 1.50(4.75) = 0 M = 7.125 kip # in.Section Properties: For the circular cross section,

A = pc2 = p(0.752) = 0.5625p in2

I =p

4c4 =

p

4(0.754) = 0.24850 in4

Maximum Normal Stress: The maximum normal stress occurs at the bottom of the circular cross section. For the combined loadings,

smax =NA

+McI

=1.50

0.5625p+

7.125(0.75)

0.24850

= 22.35 ksi = 22.4 ksi (T) Ans.

4 in.

P P

Ans:smax = 22.4 ksi (T)

754

.

*824.

The column is built up by gluing the two boards together. Determine the maximum normal stress on the cross section when the eccentric force of P = 50 kN is applied.

SolutionSection Properties: The location of the centroid of the cross section, Fig. a, is

y =yAA

=0.075(0.15)(0.3) + 0.3(0.3)(0.15)

0.15(0.3) + 0.3(0.15)= 0.1875 m

The cross-sectional area and the moment of inertia about the z axis of the cross section are

A = 0.15(0.3) + 0.3(0.15) = 0.09 m2

Iz =112

(0.3)(0.153) + 0.3(0.15)(0.1875 - 0.075)2 +112

(0.15)(0.33) + 0.15(0.3)(0.3 - 0.1875)2

= 1.5609(10- 3) m4

Equivalent Force System: Referring to Fig. b,

+ cFx = (FR)x; -50 = -F F = 50 kN

Mz = (MR)z; -50(0.2125) = -M M = 10.625 kN # mNormal Stress: The normal stress is a combination of axial and bending stress. Thus,

s =NA

+My

I

By inspection, the maximum normal stress occurs at points along the edge where y = 0.45 - 0.1875 = 0.2625 m such as point A. Thus,

smax =-50(103)

0.09-

10.625(103)(0.2625)

1.5609(10- 3)

= -2.342 MPa = 2.34 MPa (C) Ans.

150 mm

150 mm

250 mm

75 mm

300 mm

50 mm

P

Ans:smax = 2.34 MPa (C)

755

.

825.

The column is built up by gluing the two boards together. If the wood has an allowable normal stress of sallow = 6 MPa, determine the maximum allowable eccentric force P that can be applied to the column.

SolutionSection Properties: The location of the centroid c of the cross section, Fig. a, is

y =yAA

=0.075(0.15)(0.3) + 0.3(0.3)(0.15)

0.15(0.3) + 0.3(0.15)= 0.1875 m

The cross-sectional area and the moment of inertia about the z axis of the cross section are

A = 0.15(0.3) + 0.3(0.15) = 0.09 m2

Iz =112

(0.3)(0.153) + 0.3(0.15)(0.1875 - 0.075)2 +112

(0.15)(0.33) + 0.15(0.3)(0.3 - 0.1875)2

= 1.5609(10- 3) m4

Equivalent Force System: Referring to Fig. b,

+cFx = (FR)x; -P = -F F = P

Mz = (MR)z; -P(0.2125) = -M M = 0.2125P

Normal Stress: The normal stress is a combination of axial and bending stress. Thus,

F =NA

+My

I

By inspection, the maximum normal stress, which is compression, occurs at points along the edge where y = 0.45 - 0.1875 = 0.2625 m such as point A. Thus,

-6(106) =-P0.09

-0.2125P(0.2625)

1.5609(10- 3)

P = 128 076.92 N = 128 kN Ans.

150 mm

150 mm

250 mm

75 mm

300 mm

50 mm

P

Ans:Pmax = 128 kN

756

.

SolutionA = 0.75(0.5) = 0.375 in2

I =112

(0.5)(0.753) = 0.017578 in4

smax =PA

+McI

=500

0.375+

2000(0.375)

0.017578= 44.0 ksi (T) Ans.

826.

The screw of the clamp exerts a compressive force of 500 lb on the wood blocks. Determine the maximum normal stress along section aa. The cross section is rectangular, 0.75 in. by 0.50 in.

4 in.

0.75 in.a

a

Ans:smax = 44.0 ksi (T)



757

.

SolutionA = 0.75(0.5) = 0.375 in2

I =112

(0.5)(0.753) = 0.017578 in4

smax =PA

+McI

=500

0.375+

2000(0.375)

0.017578= 44.0 ksi (T)

smin =PA

-McI

=500

0.375-

2000(0.375)

0.017578= -41.3 ksi (C)

y

41.33=

(0.75 - y)44.0

y = 0.363 in.

827.

The screw of the clamp exerts a compressive force of 500 lb on the wood blocks. Sketch the stress distribution along section aa of the clamp. The cross section is rectangular, 0.75 in. by 0.50 in.

4 in.

0.75 in.a

a

Ans:smax = 44.0 ksi (T),smin = -41.3 ksi (C)

758

.

*828.

The joint is subjected to the force system shown. Sketch the normal-stress distribution acting over section aa if the member has a rectangular cross section of width 0.5 in. and thickness 1 in.

aA

Ba

1 in.2 in.

500 lb

3

45

1.5 in.

250 lb

AB

SolutionInternal Loadings: Consider the equilibrium of the lower segment of the joint sectioned through aa, Fig. a.

+S Fx = 0; V - 250 - 500 a35b = 0 V = 550 lb

+ cFy = 0; N - 500 a45 b = 0 N = 400 lb

a+Mo = 0; M + 500 a45 b(1.5) - 500 a35b(2) - 250(2) = 0

M = 500 lb # in.Section Properties: For the rectangular cross section,

A = 0.5(1) = 0.5 in2

I =112

(0.5)(13) = 0.04167 in4

Normal Stress: For the combined loadings, the normal stress at points B and C, Fig. b, can be determined from

s =NA

{ McI

=4000.5

{500(0.5)

0.04167

sC = 800 + 6000 = 6800 psi = 6.80 ksi (T) Ans.

sB = 800 - 6000 = -5200 psi = 5.20 ksi (C) Ans.

Using similar triangles,

1 - y5.20

=y

6.80 ; y = 0.5667 in.

Using these results, the normal stress distribution over section aa, shown in Fig. b, can be sketched.

Ans: sC = 6.80 ksi (T),sB = 5.20 ksi (C)

759

.

829.

The joint is subjected to the force system shown. Determine the state of stress at points A and B, and sketch the results on differential elements located at these points. The member has a rectangular cross-sectional area of width 0.5 in. and thickness 1 in.

aA

Ba

1 in.2 in.

500 lb

3

45

1.5 in.

250 lb

AB

SolutionInternal Loadings: Consider the equilibrium of the lower segment of the joint sectioned through aa, Fig. a,

+S Fx = 0; V - 250 - 500 a35b = 0 V = 550 lb

+ cFy = 0; N - 500 a45 b = 0 N = 400 lb

a+Mo = 0; M + 500 a45 b(1.5) - 500 a35b(2) - 250(2) = 0

M = 500 lb # in.Section Properties: For the rectangular cross section,

A = 0.5(1) = 0.5 in2

I =112

(0.5)(13) = 0.04167 in4

For points A and B shown in Fig. b,

QB = 0 QA = yA = 0.25[0.5(0.5)] = 0.0625 in3

Normal Stress: For the combined loadings, the normal stress at points A and B, Fig. b, can be determined from

sA =NA

+MyA

I=

4000.5

+500(0)

0.04167= 800 psi = 0.800 ksi (T) Ans.

sB =NA

-McI

=4000.5

-500(0.5)

0.04167= - 5200 psi = 5.20 ksi (C) Ans.

Shear Stress: Applying the shear formula,

tA =VQA

It=

550(0.0625)

0.04167(0.5)= 1650 psi = 1.65 ksi Ans.

tB =VQB

It= 0 Ans.

Using these results, the state of stress acting on the differential elements at points A and B shown in Fig. c can be sketched.

Ans:sA = 0.800 ksi (T),sB = 5.20 ksi (C),tA = 1.65 ksi,tB = 0

760

.

Solution

Support Reactions: Referring to the free-body diagram of the handle shown in Fig. a,

a+MD = 0; 100(0.25) - FC (0.05) = 0 FC = 500 N

Internal Loadings: Consider the equilibrium of the free-body diagram of the segment shown in Fig. b.

Fy = 0; 500 - V = 0 V = 500 Na+MC = 0; M - 500(0.025) = 0 M = 12.5 N # m

Section Properties: The moment of inertia of the cross section about the centroidal axis is

I =112

(0.0075)(0.023) = 5(10- 9) m4

Referring to Fig. c, QA and QB are

QA = 0

QB = yA = 0.005(0.01)(0.0075) = 0.375(10- 6) m3

Normal Stress: The normal stress is contributed by bending stress only. Thus

s =My

I

For point A, y = 0.01 m. Then

sA = - 12.5(0.01)

5(10- 9)= -25 MPa = 25 MPa (C) Ans.

For point B, y = 0. Then

sB = 0 Ans.

830.

The rib-joint pliers are used to grip the smooth pipeC. If the force of 100 N is applied to the handles, determine the state of stress at points A and B on the cross section of the jaw at section aa. Indicate the results on an element at each point.

250 mm

100 N

100 N

a

Section a a

a

A

B

C

25 mm25 mm

10 mm

20 mm 7.5 mm

45

761

.

830. Continued

Ans:sA = 25 MPa (C), sB = 0,tA = 0, tB = 5 MPa

Shear Stress: The shear stress is contributed by transverse shear stress only. Thus,

tA =VQA

It= 0 Ans.

tB =VQB

It=

500[0.375(10- 6)]

5(10- 9)(0.0075)= 5 MPa Ans.

The state of stress of points A and B are represented by the elements shown in Figs. d and e respectively.

762

.

Solution+S Fx = 0; NA - 150 cos 30 = 0

NA = 129.9038 lb

+ cFy = 0; VA - 150 sin 30 = 0

VA = 75 lb

a+MA = 0; 150 cos 30(1.5) + 150 sin 30(2) - MA = 0

MA = 344.8557 lb # in.

sA =PA

+McI

=129.9038

p(14)2+

344.8557(14)p4 (

14)4

= 28.8 ksi Ans.

tA = 0 (Since QA = 0) Ans.

831.

The 12-in.-diameter bolt hook is subjected to the load of F = 150 lb. Determine the stress components at point A on the shank. Show the result on a volume element located at this point.

2 in.

1.5 in.30

F = 150 lb

A

B2 in.

Ans: sA = 28.8 ksi,tA = 0

763

.

Solution+S Fx = 0; NB - 150 cos 30 = 0; NB = 129.9038

+ cFy = 0; VB - 150 sin 30 = 0; VB = 75 lb

a+MB = 0; 150 cos 30(1.5) + 150 sin 30(4) - MB = 0

MB = 494.8557 lb # in.

sB =PA

-McI

=129.9038

p(14)2-

494.8557(14)p4 (

14)4

= -39.7 ksi Ans.

*832.

The 12-in.-diameter bolt hook is subjected to the load of F = 150 lb. Determine the stress components at point B on the shank. Show the result on a volume element located at this point.

2 in.

1.5 in.30

F = 150 lb

A

B2 in.

Ans: sB = -39.7 ksi

764

.

833.

The block is subjected to the eccentric load shown. Determine the normal stress developed at points A and B. Neglect the weight of the block.

A B

C

a

a

100 mm150 kN

150 mm

SolutionInternal Loadings: Consider the equilibrium of the upper segment of the block sectioned through aa, Fig. a.

Fx = 0; N + 150 = 0 N = -150 kN

My = 0; My + 150(0.05) = 0 My = -7.50 kN # mMz = 0; Mz - 150(0.075) = 0 Mz = 11.25 kN # m


A = 0.1(0.15) = 0.015 m2

Iy =112

(0.15)(0.13) = 12.5(10-6) m4

Iz =112

(0.1)(0.153) = 28.125(10-6) m4

Normal Stresses: For the combined loadings, the normal stress can be determined from

s =NA

-Mzy

Iz+

Myz

Iy

For point A, yA = 0.075 m and zA = 0.05 m

sA =-150(103)

0.015-

11.25(103)(0.075)28.125(10-6)

+3 -7.50(103) 4(0.05)

12.5(10-6)

= -70.0(106) Pa = 70.0 MPa (C) Ans.

For point B, yB = 0.075 m and zB = -0.05 m

sB =-150(103)

0.015-

11.25(103)(0.075)28.125(10-6)

+3 -7.50(103) 4(-0.05)

12.5(10-6)

= -10.0(106) Pa = 10.0 MPa (C) Ans.

Ans:sA = 70.0 MPa (C),sB = 10.0 MPa (C)

765

.

834.

The block is subjected to the eccentric load shown. Sketch the normal-stress distribution acting over the cross section at section aa. Neglect the weight of the block.

A B

C

a

a

100 mm150 kN

150 mm

SolutionInternal Loadings: Consider the equilibrium of the upper segment of the block sectioned through aa, Fig. a.

Fx = 0; N + 150 = 0 N = -150 kN

My = 0; My + 150(0.05) = 0 My = -7.50 kN # mMz = 0; Mz - 150(0.075) = 0 Mz = 11.25 kN # m


A = 0.1(0.15) = 0.015 m2

Iy =112

(0.15)(0.13) = 12.5(10-6) m4

Iz =112

(0.1)(0.153) = 28.125(10-6) m4

Normal Stress: For the combined loadings, the normal stress can be determined from

s =NA

-Mzy

Iz+

Myz

Iy

For point A, yA = 0.075 m and zA = 0.05 m

sA =-150(103)

0.015-

11.25(103)(0.075)28.125(10-6)

+(-7.50)(103)(0.05)

12.5(10-6)

= -70.0(106) Pa = 70.0 MPa (C) Ans.

For point B, yB = 0.075 m and zB = -0.05 m

sB =-150(103)

0.015-

11.25(103)(0.075)28.125(10-6)

+(-7.50)(103)(-0.05)

12.5(10-6)

= -10.0(106) Pa = 10.0 MPa (C) Ans.

For point C, yC = -0.075 m and zC = -0.05 m

sC =-150(103)

0.015-

11.25(103)(-0.075)28.125(10-6)

+(-7.50)(103)(-0.05)

12.5(10-6)

= 50.0(106) Pa = 50.0 MPa (T) Ans.

For point D, yD = -0.075 m and zD = 0.05 m

sD =-150(103)

0.015-

11.25(103)(-0.075)28.125(10-6)

+(-7.50)(103)(0.05)

12.5(10-6)

= -10.0(106) Pa = 10.0 MPa (C) Ans.

766

.

834. Continued

The location of neutral axis can be found using similar triangles.

y

10.0=

0.15 - y50.0

; y = 0.025 m

z10.0

=0.1 - z

50.0; z = 0.01667 m

Using these result, the normal stress distribution over the cross section shown in Fig. b can be sketched.

Ans:sA = 70.0 MPa (C),sB = 10.0 MPa (C),sC = 50.0 MPa (T),sD = 10.0 MPa (C)

767

.

Ans:sA = 27.3 ksi (T),sB = 0.289 ksi (T),tA = 0,tB = 0.750 ksi

SolutionSupport Reactions:

+ cFy = 0; 2000 - 2F cos 30 = 0 F = 1154.70 lb

Internal Forces and Moment:

+> Fx = 0; 1154.70 sin 30 - N = 0 N = 577.35 lb

+ cFy = 0; V - 1154.70 cos 30 = 0 V = 1000 lb

a+MB = 0; M - 1154.70 cos 30(1.5) = 0

M = 1500 lb # ftSection Properties:

A = 1(2) = 2.00 in2

I =112

(1)(23) = 0.6666 in4

QB = yA = 0.5(1)(1) = 0.500 in3

QA = 0

Normal Stress:

s =NA

{My

I

sA =577.352.00

+1500(12)(1)

0.6666

= 27 300 psi = 27.3 ksi (T) Ans.

sB =577.352.00

+1500(12)(0)

0.6666

= 289 psi = 0.289 ksi (T) Ans.


t =VQ

It

tA = 0 Ans.

tB =1000(0.500)

0.6666(1)

= 750 psi = 0.750 ksi Ans.

835.

The spreader bar is used to lift the 2000-lb tank. Determine the state of stress at points A and B, and indicate the results on a differential volume element.

1.5 ft0.5 ft3030

A

B

1 in.1 in.

AB

768

.

SolutionInternal Loadings: Consider the equilibrium of the free-body diagram of the drills right cut segment, Fig. a.

Fx = 0; N - 150 a45b = 0 N = 120 N

Fy = 0; 150 a35 b - Vy = 0 Vy = 90 N

Mx = 0; 20 - T = 0 T = 20 N # m

Mz = 0; -150 a35 b(0.4) + 150 a45b(0.125) + Mz = 0

Mz = 21 N # mSection Properties: The cross-sectional area, the moment of inertia about the z axis, and the polar moment of inertia of the drills cross section are

A = p10.00522 = 25p110- 62 m2

Iz =p

4 10.00542 = 0.15625p110- 92 m4

J =p

2 10.00542 = 0.3125p110- 92 m4

Referring to Fig. b, QA is

QA = 0


s =NA

-Mzy

Iz

For point A, y = 0.005 m. Then

sA =-120

25p110- 62 -21(0.005)

0.15625p110- 92 = -215.43 MPa = 215 MPa (C) Ans.

*836.

The drill is jammed in the wall and is subjected to the torque and force shown. Determine the state of stress at point A on the cross section of the drill bit at section aa.

150 N

34

5

125 mm

20 Nm

400 mm

a

a

5 mm

B

A

Section a a

z

x

y

y

769

.

*836. Continued

Shear Stress: The transverse shear stress developed at point A is

c 1txy2V dA=

VyQAIzt

= 0

The torsional shear stress developed at point A is

3(txz)T4A = TcJ =20(0.005)

0.3125p110- 92 = 101.86 MPa

Thus,

1txy2A = 0 Ans.

1txz2A = c 1txz2T dA= 102 MPa Ans.

The state of stress at point A is represented on the element shown in Fig. c.

Ans:sA = 215 MPa (C),1txy2A = 0,1txz2A = 102 MPa

770

.

SolutionInternal Loadings: Consider the equilibrium of the free-body diagram of the drills right cut segment, Fig. a.

Fx = 0; N - 150a45b = 0 N = 120 N

Fy = 0; 150a35 b - Vy = 0 Vy = 90 N

Mx = 0; 20 - T = 0 T = 20 N # m

Mz = 0; -150a35 b(0.4) + 150a45b(0.125) + Mz = 0

Mz = 21 N # m

Section Properties: The cross-sectional area, the moment of inertia about the z axis, and the polar moment of inertia of the drills cross section are

A = p10.00522 = 25p110- 62 m2

Iz =p

210.00542 = 0.15625p110- 92 m4

J =p

210.00542 = 0.3125p110- 92 m4

Referring to Fig. b, QB is

QB = yA =4(0.005)

3pcp

210.00522 d = 83.333110- 92 m3


s =NA

-Mzy

Iz

For point B, y = 0. Then

sB =-120

25p110- 62 - 0 = -1.528 MPa = 1.53 MPa (C) Ans.

837.

The drill is jammed in the wall and is subjected to the torque and force shown. Determine the state of stress at point B on the cross section of the drill bit at section aa.

150 N

34

5

125 mm

20 Nm

400 mm

a

a

5 mm

B

A

Section a a

z

x

y

y

771

.

Shear Stress: The transverse shear stress developed at point B is

c 1txy2V dB=

VyQBIzt

=90 c 83.333110- 92 d

0.15625p110- 92(0.01) = 1.528 MPa

The torsional shear stress developed at point B is

c 1txy2T dB=

TcJ

=20(0.005)

0.3125p110- 92 = 101.86 MPa

Thus,

1txz2B = 0 Ans. 1txy2B = c 1txy2T d

B- c 1txy2V d

B

= 101.86 - 1.528 = 100.33 MPa = 100 MPa Ans.

The state of stress at point B is represented on the element shown in Fig. d.

837. Continued

Ans:sB = 1.53 MPa (C), 1txz2B = 100 MPa

772

.

Solution

sD = -PA

-My

I= -

8(103)

(0.1)(0.05)-

12(103)(0.03)112 (0.05)(0.1)

3

sD = -88.0 MPa Ans.

tD = 0 Ans.

838.

The frame supports the distributed load shown. Determine the state of stress acting at point D. Show the results on a differential element at this point.

4 kN/m

DB

A

C

E

1.5 m 1.5 m

20 mm

50 mm

20 mm

60 mm

3 m

3 m

5 m

D

E

Ans:sD = -88.0 MPa, tD = 0

773

.

Solution

sE = -PA

-My

I=

8(103)

(0.1)(0.05)+

8.25(103)(0.03)112(0.05)(0.1)

3= 57.8 MPa Ans.

tE =VQ

I t=

4.5(103)(0.04)(0.02)(0.05)112 (0.05)(0.1)

3(0.05)= 864 kPa Ans.

839.

The frame supports the distributed load shown. Determine the state of stress acting at point E. Show the results on a differential element at this point.

4 kN/m

DB

A

C

E

1.5 m 1.5 m

20 mm

50 mm

20 mm

60 mm

3 m

3 m

5 m

D

E

Ans:sE = 57.8 MPa, tE = 864 kPa

774

.

*840.

The rod has a diameter of 40 mm. If it is subjected to the force system shown, determine the stress components that act at point A, and show the results on a volume element located at this point.

SolutionInternal Loadings: Consider the equilibrium of the left segment of the rod being sectioned, Fig. a.

Fx = 0; Nx - 1500 = 0 Nx = 1500 N

Fy = 0; Vy - 600 = 0 Vy = 600 N

Fz = 0; Vz + 800 = 0 Vz = -800 N

Mx = 0; Tx - 100 = 0 Tx = 100 N # mMy = 0; My + 800(0.3) = 0 My = -240 N # mMz = 0; Mz + 600(0.3) = 0 Mz = -180 N # m

Section Properties: For the circular cross section, Fig. b,

A = pc2 = p(0.022) = 0.400(10-3)p m2

Iy = Iz =p

4c4 =

p

4(0.024) = 40.0(10-9)p m4

J =p

2c4 =

p

2(0.024) = 80.0(10-9)p m4

(QA)z = 0 (QA)y = yA =4(0.02)

3pcp

2(0.022) d = 5.3333(10-6) m3

Normal Stress: For the combined loadings, the normal stress at point A can be determined from

sx = sA =NxA

-MzyA

Iz+

MyzAIy

=1500

0.400(10-3)p-

(-180)(0)40.0(10-9)p

+(-240)(0.02)40.0(10-9)p

= -37.00(106) Pa = 37.0 MPa (C) Ans.

Shear Stress: The transverse shear stress in z and y directions and the torsional shear

stress can be obtained using the shear formula tV =VQ

It and the torsion formula

tT =Tr

J respectively.

(txy)A = (tV)y - tT

=60035.3333(10-6) 440.0(10-9)p(0.04)

-100(0.02)

80.0(10-9)p

= -7.321(106) Pa = -7.32 MPa Ans.

(txz)A = (tV)z = 0 Ans.

Using these results, the state of stress at point A can be represented by the differential volume element show in Fig. c.

100 mm

300 mm

y

x

BA

1500 N

800 N600 N

100 Nm

z

Ans:sA = 37.0 MPa (C),(txy)A = -7.32 MPa,(txz)A = 0

775

.

841.

Solve Prob. 840 for point B.


Fx = 0; Nx - 1500 = 0 Nx = 1500 N

Fy = 0; Vy - 600 = 0 Vy = 600 N

Fz = 0; Vz + 800 = 0 Vz = -800 N

Mx = 0; Tx - 100 = 0 Tx = 100 N # mMy = 0; My + 800(0.3) = 0 My = -240 N # mMz = 0; Mz + 600(0.3) = 0 Mz = -180 N # m

Section Properties: For the circular cross section,

A = pc2 = p(0.022) = 0.400(10-3)p m2

Iy = Iz =p

4c4 =

p

4(0.024) = 40.0(10-9)p m4

J =p

2c4 =

p

2(0.024) = 80.0(10-9)p m4

(QB)z = zA =4(0.02)

3pcp

2(0.022) d = 5.3333(10-6) m3 (QB)y = 0

Normal Stress: For the combined loadings, the normal stress at point B can be determined from

sx = sB =NxA

-MzyB

Iz+

MyzBIy

=1500

0.400(10-3)p-

(-180)(-0.02)40.0(10-9)p

+(-240)(0)

40.0(10-9)p

= -27.45(106) = 27.5 MPa (C) Ans.




tT =Tr

J, respectively.

(txz)B = (tV)z - tT

=-80035.3333(10-6)440.0(10-9)p(0.04)

-100(0.02)

80.0(10-9)p

= -8.807(106) Pa = -8.81 MPa Ans.

(txy)B = (tV)y = 0 Ans.

Using these results, the state of stress at point B, can be represented by the differential volume element shown in Fig. c.

100 mm

300 mm

y

x

BA

1500 N

800 N600 N

100 Nm

z

Ans:sB = 27.5 MPa (C),(txz)B = -8.81 MPa,(txy)B = 0

776

.

SolutionFx = 0; Vx - 125 = 0; Vx = 125 lb

Fy = 0; 75 - Ny = 0; Ny = 75 lb

Fz = 0; Vz - 200 = 0; Vz = 200 lb

Mx = 0; 200(8) - Mx = 0; Mx = 1600 lb # in.My = 0; 200(3) - Ty = 0; Ty = 600 lb # in.Mz = 0; Mz + 75(3) - 125(8) = 0; Mz = 775 lb # in.A = p(0.52) = 0.7854 in2

J =p

2(0.54) = 0.098175 in4

I =p

2(0.54) = 0.049087 in4

(QA)x = 0

(QA)z =4(0.5)

3p a1

2b(p)(0.52) = 0.08333 in3

(sA)y = -NyA

+Mxc

I

= -75

0.7854+

1600(0.5)

0.049087

= 16202 psi = 16.2 ksi (T) Ans.

(tA)yx = (tA)V - (tA)twist

=Vx(QA)Z

I t-

Tyc

J

=125(0.08333)

0.049087(1)-

600(0.5)

0.098175

= -2843 psi = -2.84 ksi Ans.

(tA)yz =Vz(QA)x

It= 0 Ans.

842.

The beveled gear is subjected to the loads shown. Determine the stress components acting on the shaft at point A, and show the results on a volume element located at this point. The shaft has a diameter of 1 in. and is fixed to the wall at C. C

B

x

zy

A

200 lb

125 lb75 lb

8 in.

3 in.

Ans:(sA)y = 16.2 ksi (T),(tA)yx = -2.84 ksi, (tA)yz = 0

777

.

SolutionFx = 0; Vx - 125 = 0; Vx = 125 lb

Fy = 0; 75 - Ny = 0; Ny = 75 lb

Fz = 0; Vz - 200 = 0; Vz = 200 lb

Mx = 0; 200(8) - Mx = 0; Mx = 1600 lb # in.My = 0; 200(3) - Ty = 0; Ty = 600 lb # in.Mz = 0; Mz + 75(3) - 125(8) = 0; Mz = 775 lb # in.A = p(0.52) = 0.7854 in2

J =p

2(0.54) = 0.098175 in4

I =p

2(0.54) = 0.049087 in4

(QB)z = 0

(QB)x =4(0.5)

3p a1

2b(p)(0.52) = 0.08333 in3

(sB)y = -PyA

+Mzc

I

= -75

0.7854+

775(0.5)

0.049087

= 7.80 ksi (T) Ans.

(tB)yz = (tB)V + (tB)twist

=Vz(QB)x

It-

Tyc

J

=200(0.08333)

0.049087 (1)+

600(0.5)

0.098175

= 3395 psi = 3.40 ksi Ans.

(tB)yx =Vx(QB)z

It= 0 Ans.

843.

The beveled gear is subjected to the loads shown. Determine the stress components acting on the shaft at point B, and show the results on a volume element located at this point. The shaft has a diameter of 1 in. and is fixed to the wall at C. C

B

x

zy

A

200 lb

125 lb75 lb

8 in.

3 in.

Ans:(sB)y = 7.80 ksi (T),(tB)yz = 3.40 ksi,(tB)yx = 0

778

.

SolutionReferring to Fig. a,

Fx = (FR)x; -6 - 12 = F F = -18.0 kip

My = (MR)y; 6(1.5) - 12(1.5) = My My = -9.00 kip # inMz = (MR)z; 12(3) - 6(3) = Mz Mz = 18.0 kip # in

The cross-sectional area and moments of inertia about the y and z axes of the cross section are

A = 6(3) = 18 in2

Iy =112

(6)(3)3 = 13.5 in4

Iz =112

(3)(63) = 54.0 in4

The normal stress developed is the combination of axial and bending stress. Thus,

s =FA

-Mz y

Iz+

My z

Iy

For point A, y = 3 in. and z = -1.5 in.

sA =-18.018.0

-18.0(3)

54.0+

-9.00(-1.5)13.5

= -1.00 ksi = 1.00 ksi (C) Ans.

For point B, y = 3 in and z = 1.5 in.

sB =-18.018.0

-18.0(3)

54+

-9.00(1.5)13.5

= -3.00 ksi = 3.00 ksi (C) Ans.

*844.

Determine the normal stress developed at points A and B. Neglect the weight of the block.

a

a

6 in.

6 kip

12 kip3 in.

AD

B

Ans:sA = 1.00 ksi (C),sB = 3.00 ksi (C)

779

.

SolutionReferring to Fig. a,

Fx = (FR)x; -6 - 12 = F F = -18.0 kip

My = (MR)y; 6(1.5) - 12(1.5) = My My = -9.00 kip # inMz = (MR)z; 12(3) - 6(3) = Mz Mz = 18.0 kip # in

The cross-sectional area and the moment of inertia about the y and z axes of the cross section are

A = 3 (6) = 18.0 in2

Iy =112

(6)(33) = 13.5 in4

Iz =112

(3)(63) = 54.0 in4

The normal stress developed is the combination of axial and bending stress. Thus,

s =FA

-Mzy

Iz+

Myz

Iy

For point A, y = 3 in. and z = -1.5 in.

sA =-18.018.0

-18.0(3)

54.0+

-9.00(-1.5)13.5

= -1.00 ksi = 1.00 ksi (C) Ans.

For point B, y = 3 in. and z = 1.5 in.

sB =-18.018.0

-18.0(3)

54.0+

-9.00(1.5)13.5

= -3.00 ksi = 3.00 ksi (C) Ans.

845.

Sketch the normal-stress distribution acting over the cross section at section aa. Neglect the weight of the block.

a

a

6 in.

6 kip

12 kip3 in.

AD

B

Ans:sA = 1.00 ksi (C),sB = 3.00 ksi (C)

780

.

For point C, y = -3 in. and z = 1.5 in.

sC =-18.018.0

-18.0(-3)

54.0+

-9.00(1.5)13.5

= -1.00 ksi = 1.00 ksi (C) Ans.

For point D, y = -3 in. and z = -1.5 in.

sD =-18.018.0

-18.0(-3)

54.0+

-9.00(-1.5)13.5

= 1.00 ksi (T) Ans.

The normal stress distribution over the cross section is shown in Fig. b

845. Continued

Ans:sA = 1.00 ksi (C), sB = 3.00 ksi (C),sC = 1.00 ksi (C), sD = 1.00 ksi (T)

781

.

SolutionEquilibrium: For the man

a+MB = 0; 981(1) - 2FA(1.35) = 0 FA = 363.33 N

Section Properties:

r- = 0.15 +0.025

2= 0.1625 m

LA dAr

= 2p( r - 2r 2 - c2)= 2p(0.1625 - 20.16252 - 0.01252)= 3.02524(10- 3) m

A = p(0.01252) = 0.490874(10-3) m2

R =A

1A dAr=

0.490874(10-3)3.02524(10-3)

= 0.162259 m

r- - R = 0.1625 - 0.162259 = 0.240741(10-3) m

Internal Force and Moment: The internal moment must be computed about the neutral axis.

+ cFy = 0; -363.33 - N = 0 N = -363.33 N

a+Mo = 0; -M - 363.33(0.462259) = 0 M = -167.95 N # m

Normal Stress: Apply the curved-beam formula.For tensile stress

(st)max =NA

+M(R - r2)Ar2(r

- - R)

=-363.33

0.490874(10-3)+

-167.95(0.162259 - 0.175)0.490874(10-3)(0.175)0.240741(10-3)

= 103 MPa (T) Ans.For compressive stress,

(sc)max =NA

+M(R - r1)Ar1(r

- - R)

=-363.33

0.490874(10-3)+

-167.95(0.162259 - 0.15)0.490874(10-3)(0.15)0.240741(10-3)

= -117 MPa = 117 MPa (C) Ans.

846.

The man has a mass of 100 kg and center of mass at G. If he holds himself in the position shown, determine the maximum tensile and compressive stress developed in the curved bar at section aa. He is supported uniformly by two bars, each having a diameter of 25 mm. Assume the floor is smooth. Use the curved-beam formula to calculate the bending stress.

G

aa

300 mm

150 mm

0.35 m 1 m

300 mm

Ans:(st)max = 103 MPa (T),(sc)max = 117 MPa (C)

782

.

847.

The solid rod is subjected to the loading shown. Determine the state of stress at point A, and show the results on a differential volume element located at this point.


Fx = 0; Nx - 100 = 0 Nx = 100 kN

Fy = 0; Vy - 10 = 0 Vy = 10 kN

Fz = 0; Vz + 20 = 0 Vz = -20 kN

Mx = 0; Tx + 20(0.03) + 10(0.03) = 0 Tx = -0.9 kN # mMy = 0; My + 20(0.2) + 100(0.03) = 0 My = -7.00 kN # mMz = 0; Mz + 10(0.4) = 0 Mz = -4.00 kN # m

Section Properties: For the circular cross section, Fig. b,A = pc2 = p(0.032) = 0.9(10-3)p m2

Iy = Iz =p

4c4 =

p

4(0.034) = 0.2025(10-6)p m4

J =p

2c4 =

p

2(0.034) = 0.405(10-6)p m4

(QA)z = zA =4(0.03)

3pcp

2(0.032) d = 18.0(10-6) m3

(QA)y = 0


sx = sA =NxA

-MzyA

Iz+

MyzAIy

=100(103)

0.9(10-3)p-

3 -4.00(103) 4(0.03)0.2025(10-6)p

+3 -7.00(103) 4(0)0.2025(10-6)p

= 224.00(106) Pa = 224 MPa (T) Ans.




tT =Tr

J, respectively.

(txz)A = (tV)z + tT

=-20(103) 318.0(10-6) 40.2025(10-6)p(0.06)

+-0.9(103)(0.03)0.405(10-6)p

= -30.65(106) Pa = -30.7 MPa Ans.

(txy)A = (tV)y = 0 Ans.

Using these results, the state of stress at point A can be represented by the differential volume element shown in Fig. c.

Ans:sA = 224 MPa (T),(txz)A = -30.7 MPa,(txy)A = 0

30 mm

Ax

y

zB

C

100 kN

10 kN

20 kN

200 mm200 mm

783

.

*848.

The solid rod is subjected to the loading shown. Determine the state of stress at point B, and show the results on a differential volume element at this point.


Fx = 0; Nx - 100 = 0 Nx = 100 kN

Fy = 0; Vy - 10 = 0 Vy = 10 kN

Fz = 0; Vz + 20 = 0 Vz = -20 kN



A = pc2 = p(0.032) = 0.9(10-3)p m2

Iy = Iz =p

4c4 =

p

4(0.034) = 0.2025(10-6)p m4

J =p

2c4 =

p

2(0.034) = 0.405(10-6)p m4

(QB)z = zA =4(0.03)

3pcp

2(0.032) d = 18.0(10-6) m3

(QB)y = 0


sx = sB =NxA

-MzyB

Iz+

MyzBIy

=100(103)

0.9(10-3)p-

3 -4.00(103) 4(-0.03)0.2025(10-6)p

+3 -7.00(103) 4(0)0.2025(10-6)p

= -153.26(106) Pa = 153 MPa (C) Ans.


stress can be determined using the shear formula tV =VQ


tT =Tr

J, respectively.

(txz)B = -tT + (tV)z

=0.9(103)(0.03)0.405(10-6)p

+-20(103) 318.0(10-6) 40.2025(10-6)p(0.06)

= 11.79(106) Pa = 11.8 MPa Ans.

(txy)B = (tV)y = 0 Ans.

Using these results, the state of stress at point B can be represented by the differential volume element shown in Fig. c.

Ans:sB = 153 MPa (C),(txz)B = 11.8 MPa,(txy)B = 0

30 mm

Ax

y

zB

C

100 kN

10 kN

20 kN

200 mm200 mm

784

.

849.

The solid rod is subjected to the loading shown. Determine the state of stress at point C, and show the results on a differential volume element at this point.


Fx = 0; Nx - 100 = 0 Nx = 100 kN

Fy = 0; Vy - 10 = 0 Vy = 10 kN

Fz = 0; Vz + 20 = 0 Vz = -20 kN



A = pc2 = p(0.032) = 0.9(10-3)p m2

Iy = Iz =p

4c4 =

p

4(0.034) = 0.2025(10-6)p m4

J =p

2c4 =

p

2(0.034) = 0.405(10-6)p m4

(QC)y = yA = c4(0.03)

3pd cp

2(0.032) d = 18.0(10-6) m3

(QC)z = 0

Normal Stress: For the combine loadings, the normal stress at point C can be determined from

sx = sC =NxA

-MzyC

Iz+

MyzCIy

=100(103)

0.9(10-3)p-

3 -4.00(103) 4(0)0.2025(10-6)p

+3 -7.00(103) 4(0.03)

0.2025(10-6)p

= -294.73(106) Pa = 295 MPa (C) Ans.


stress can be determined using the shear formula tV =VQ


tT =Tr

J, respectively.

(txy)C = (tV)y - tT

=10(103) 318.0(10-6) 40.2025(10-6)p(0.06)

--0.9(103)(0.03)0.405(10-6)p

= 25.94(106) Pa = 25.9 MPa Ans.

(txz)C = (tV)z = 0 Ans.

Using these results, the state of stress at point C can be represented by the differential volume element shown in Fig. c.

Ans:sC = 295 MPa (C),(txy)C = 25.9 MPa,(txz)C = 0

30 mm

Ax

y

zB

C

100 kN

10 kN

20 kN

200 mm200 mm

785

.

SolutionRequire sA = 0

sA = 0 =PA

+McI

; 0 =-Ppc2

+(Pe)c

p

4 c4

e =c4

Ans.

850.

The post has a circular cross section of radius c. Determine the maximum radius e at which the load P can be applied so that no part of the post experiences a tensile stress. Neglect the weight of the post.

P

c

e

Ans:

e =c4

786

.

SolutionEquivalent Force System: As shown on FBD.

Section Properties:

A = 2a(2a) + 2J12

(2a)aR = 6a2 Iz =

112

(2a)(2a)3 + 2J 136

(2a) a3 +12

(2a) aaa + a3b

2R = 5a4

Iy =112

(2a)(2a)3 + 2J 136

(2a) a3 +12

(2a) aaa3b

2R =

53

a4

Normal Stress:

s =NA

-Mzy

Iz+

My z

Iy

=-P6a2

-Peyy

5a4 -

Pez z53 a

4

=P

30a4 1 -5a2 - 6eyy - 18ez z2

At point B where y = -a and z = -a, we require sB 6 0.

0 7P

30a4 3 -5a2 - 6(-a) ey - 18(-a) ez4

0 7 -5a + 6ey + 18ez

6ey + 18ez 6 5a Ans.

When ez = 0, ey 656

a

When ey = 0, ez 6518

a

Repeat the same procedures for point A, C and D. The region where P can be applied without creating tensile stress at points A, B, C, and D is shown shaded in the diagram.

851.

The post having the dimensions shown is subjected to the load P. Specify the region to which this load can be applied without causing tensile stress at points A, B, C, and D.

x

y

z

A

a a a a

a

a

Dez

eyBC

P

Ans:6ey + 18ez 6 5a

787

.

Ans:

P =dmaxp(r 0

4 - r i4)r 0

2 + r i2 + 4er0

Solution

smax =PA

+Pero

p

4 (r 0

4 - r 4i )

smax = P c 1p(r 20 - r 2i )

+4er0

p(r 04 - r 4i )

d

smax =P

p(r 02 - r i2)

c 1 + 4er0(r 0

2 + r i2)d

smax =P(r 0

2 + r i2 + 4er0)p(r 0

2 - r i2)(r 02 + r i2)

smax =P(r 0

2 + r i2 + 4er0)p(r 0

4 - r i4)

P =dmaxp(r 0

4 - r i4)r 0

2 + r i2 + 4er0 Ans.

*852.

The vertebra of the spinal column can support a maximum compressive stress of smax, before undergoing a compression fracture. Determine the smallest force P that can be applied to a vertebra, if we assume this load is applied at an eccentric distance e from the centerline of the bone, and the bone remains elastic. Model the vertebra as a hollow cylinder with an inner radius ri and outer radius ro.

e

P

ro

ri

788

.

Solution

TcJ

= max on perimeter =PRr

J

tmax =VA

+TcJ

=PA

+PRr

J QED

853.

The coiled spring is subjected to a force P. If we assume the shear stress caused by the shear force at any vertical section of the coil wire to be uniform, show that the maximum shear stress in the coil is tmax = P>A + PRr>J, where J is the polar moment of inertia of the coil wire and Ais its cross-sectional area.

P

P

R

r2

Ans:N/A

789

.

SolutionMember CD:

a+MC = 0; 35

FDE(16) - 28.8(8) = 0;

FDE = 24.0 kip

Segment:

+d Fx = 0 ; N -45

(24.0) = 0; N = 19.2 kip

+ cFy = 0; V +35

(24.0) - 19.8 = 0; V = 5.40 kip

a+Mo = 0; -M - 19.8(5.5) +35

(24.0)(11) = 0;

M = 49.5 kip # ft

A = 7(1.5) + 6(1) = 16.5 in2

y- =y-AA

=0.75(1.5)(7) + 4.5(6)(1)

16.5= 2.1136 in.

I =112

(7)(1.53) + 7(1.5)(2.1136 - 0.75)2 +112

(1)(63) + 1(6)(4.5 - 2.1136)2

= 73.662 in4

QA = QB = 0

Normal Stress:

s =NA

{My

I

sA =19.216.5

-49.5(12)(7.5 - 2.1136)

73.662= -15.9 ksi = 15.9 ksi (C) Ans.

sB =19.216.5

+49.5(12)(5.3864)

73.662= 44.6 ksi (T) Ans.

Shear Stress: Since QA = QB = 0,

tA = tB = 0 Ans.

854.

The frame supports a centrally applied distributed load of 1.8 kip>ft. Determine the state of stress at points A and B on member CD and indicate the results on a volume element located at each of these points. The pins at C and D are at the same location as the neutral axis for the crosssection.

12 ft

5 ft

16 ft

A

BD

C

1.8 kip/ft

1.5 in.

6 in.

1 in.3 in. 3 in.

E

A

B

Ans:sA = 15.9 ksi (C),sB = 44.6 ksi (T),tA = tB = 0

790

.

SolutionInternal Loadings: Consider the equilibrium of the right segment of the rod being sectioned, Fig. a.

Fx = 0; Nx - 1200 = 0 Nx = 1200 lb

Fy = 0; Vy + 200 = 0 Vy = -200 lb

Fz = 0; Vz + 300 = 0 Vz = -300 lb

Mx = 0; Tx + 300(3) = 0 Tx = -900 lb # in.My = 0; My + 300(9) = 0 My = -2700 lb # in.Mz = 0; Mz + 1200(3) - 200(9) = 0 Mz = -1800 lb # in.


A = pc2 = p(0.52) = 0.25p in.2

Iy = Iz =p

4(0.54) = 0.015625p in.4

J =p

2(0.54) = 0.03125p in4

(QA)z = z9=A= =

4(0.5)

3pcp

2(0.52) d = 0.08333 in3

(QA)y = 0


sx = sA =NxA

-MzyA

Iz+

MyzAIy

=12000.25p

-(-1800)(-0.5)

0.015625p+

(-2700)(0)0.015625p

= -16.81(103) psi = 16.8 ksi (C) Ans.

Shear Stress: The transverse shear stress in z and y directions and the torsional

shear stress can be obtained using the shear formula tV =VQ

It and torsion formula

tT =Tr

J, respectively.

(txz)A = -tT + (tV)z

= --900(0.5)0.03125p

+-300(0.08333)0.015625p (1)

= 4.074(103) psi = 4.07 ksi Ans.(txy)A = (tV)y = 0 Ans.

Using these results, the state of stress at point A can be represented by the differential volume element shown in Fig. c.

855.

The 1-in.-diameter rod is subjected to the loads shown. Determine the state of stress at point A, and show the results on a differential volume element located at thispoint.

B

A

y

z

x 200 lb9 in.

3 in.

300 lb

1200 lb

Ans:sA = 16.8 ksi (C),(txz)A = 4.07 ksi,(txy)A = 0

791

.

SolutionInternal Loadings: Consider the equilibrium of the right segment of the rod being sectioned, Fig. a.

Fx = 0; Nx - 1200 = 0 Nx = 1200 lb

Fy = 0; Vy + 200 = 0 Vy = -200 lb

Fz = 0; Vz + 300 = 0 Vz = -300 lb

Mx = 0; Tx + 300(3) = 0 Tx = -900 lb # in.My = 0; My + 300(9) = 0 My = -2700 lb # in.Mz = 0; Mz + 1200(3) - 200(9) = 0 Mz = -1800 lb # in.


A = pc2 = p(0.52) = 0.25p in2

Iy = Iz =p

4(0.54) = 0.015625p in4

J =p

2(0.54) = 0.03125p in4

(QB)y = y9 =A= =4(0.5)

3pcp

2 (0.52) d = 0.08333 in3

(QB)z = 0


sx = sB =NxA

-MzyB

Iz+

MyzBIy

=12000.25p

-(-1800)(0)0.015625p

+(-2700)(0.5)

0.015625p

= -25.97(103) psi = 26.0 ksi (C) Ans.

Shear Stress: The transverse shear stress in z and y directions and the torsional

shear stress can be obtained using the shear formula tV =VQ

It and torsion formula

tT =Tr

J, respectively.

(txy)B = -tT + (tV)y

=900(0.5)

0.03125p+

-200(0.08333)0.015625p (1)

= 4.244(103) psi = 4.24 ksi Ans.(txz)B = (tV)z = 0 Ans.

Using these results, the state of stress at point B can be represented by the differential volume element shown in Fig. c.

*856.

The 1-in.-diameter rod is subjected to the loads shown. Determine the state of stress at point B, and show the results on a differential volume element located at thispoint.

B

A

y

z

x 200 lb9 in.

3 in.

300 lb

1200 lb

Ans:sB = 26.0 ksi (C),(txy)B = 4.24 ksi,(txz)B = 0

792

.

SolutionPoint A:

sA =McI

=10.5(103)(0.05)

p4(0.05)

4 = 107 MPa (T) Ans.

tA =TcJ

=3(103)(0.05)

p4(0.05)

4 = 15.279(106) = 15.3 MPa Ans.

Point B:

sB = 0 Ans.

tB =TcJ

-VQ

It= 15.279(106) -

3000(4(0.05)/3p))(12)(p)(0.05)2

p4(0.05)

4(0.1)

tB = 14.8 MPa Ans.

857.

The sign is subjected to the uniform wind loading. Determine the stress components at points A and B on the 100-mm-diameter supporting post. Show the results on a volume element located at each of these points.

D

y

x

CB

A

1 m

1.5 kPa3 m

2 m

2 m

z

Ans:sA = 107 MPa (T), tA = 15.3 MPa,sB = 0, tB = 14.8 MPa

793

.

SolutionPoint C:

sC =McI

=10.5(103)(0.05)

p4(0.05)

4 = 107 MPa (C) Ans.

tC =TCJ

=3(103)(0.05)

p2(0.05)

4 = 15.279(106) = 15.3 MPa Ans.

Point D:

sD = 0 Ans.

tD =TcJ

+VQ

It= 15.279(106) +

3(103)(4(0.05)/3p)(12)(p)(0.05)2

p4(0.05)

4(0.1)= 15.8 MPa

Ans.

858.

The sign is subjected to the uniform wind loading. Determine the stress components at points C and D on the 100-mm-diameter supporting post. Show the results on a volume element located at each of these points.

D

y

x

CB

A

1 m

1.5 kPa3 m

2 m

2 m

z

Ans:sC = 107 MPa (C), tC = 15.3 MPa,sD = 0, tD = 15.8 MPa

794

.

859.

The bearing pin supports the load of 900 lb. Determine the stress components in the support member at point A. Represent the state of stress at point A with a differential element.

30

3 in.

A

AB

B

4 in.

1 in.900 lb

0.25 in.0.5 in.

0.25 in.

SolutionInternal Loadings: Consider the equilibrium of the right segment of the supports being sectioned, Fig. a.

a+Fx = 0; N - 900 cos 30 = 0 N = 779.42 lb+Q Fy = 0; 900 sin 30 - V = 0 V = 450 lb

a+Mo = 0; 900(3 sin 30 - 1) - M = 0 M = 450 lb # inSection Properties: For the square cross section,

A = 0.5(0.5) = 0.25 in2

I =112

(0.5)(0.53) = 0.0052083 in4

QA = 0

Normal Stress: For the combined loadings, the normal stress can be determined from

sA =NA

+McI

=779.420.25

+450(0.25)

0.0052083= 24.72(103) psi = 24.7 ksi (T) Ans.


tA =VQA

It= 0 Ans.

Using these results, the state of stress at point A can be represented by the differential element in Fig. b.

Ans:sA = 24.7 ksi (T),tA = 0

795

.

*860.

The bearing pin supports the load of 900 lb. Determine the stress components in the support member at point B. Represent the state of stress at point B with a differential element.

30

3 in.

A

AB

B

4 in.

1 in.900 lb

0.25 in.0.5 in.

0.25 in.

SolutionInternal Loadings: Consider the equilibrium of the right segment of the support being sectioned, Fig. a.

a+Fx = 0; N - 900 cos 30 = 0 N = 779.42 lb+Q Fy = 0; 900 sin 30 - V = 0 V = 450 lb

a+Mo = 0; 900(3 sin 30 - 1) - M = 0 M = 450 lb # inSection Properties: For the square cross section,

A = 0.5(0.5) = 0.25 in2

I =112

(0.5)(0.53) = 0.0052083 in4

For the area shown shaded in Fig. b,

QB = y- =A= = 0.12530.25(0.5)4 = 0.015625 in3Normal Stress: For the combined loadings, the normal stress can be determined from

sB =NA

+My

I=

779.420.25

+ 0 = 3.1177(103) psi = 3.12 ksi (T) Ans.


tB =VQB

It=

450(0.015625)

0.0052083(0.5)= 2.700(103) psi = 2.70 ksi Ans.

Using these results, the state of stress at point B can be represented by the differential element in Fig. c.

Ans:sB = 3.12 ksi (T),tB = 2.70 ksi

796

.

Solution

x =x-AA

=(0.005)(0.04)(0.01) + 0.04(0.06)(0.01)

0.04(0.01) + 0.06(0.01)= 0.026 m

A = 0.04(0.01) + 0.06(0.01) = 0.001 m2

I =112

(0.04)(0.013) + (0.04)(0.01)(0.026 - 0.005)2

+112

(0.01)(0.063) + 0.01(0.06)(0.040 - 0.026)2 = 0.4773(10- 6) m4

(smax)t =PA

+Mx

I=

8(103)

0.001+

1.808(103)0.026

0.4773(10- 6)

= 106.48 MPa = 106 MPa Ans.

(smax)c =PA

-McI

=8(103)

0.001-

1.808(103)(0.070 - 0.026)0.4773(10- 6)

= -158.66 MPa = -159 MPa Ans.

x158.66

=70 - x106.48

; x = 41.9 mm

861.

The C-frame is used in a riveting machine. If the force at the ram on the clamp at D is P = 8 kN, sketch the stress distribution acting over the section aa.

D

a

40 mm

10 mm

60 mm 10 mm

200 mm

a P

Ans:(smax)t = 106 MPa, (smax)c = -159 MPa

797

.

Solution

x =xAA

=(0.005)(0.04)(0.01) + 0.04(0.06)(0.01)

0.04(0.01) + 0.06(0.01)= 0.026 m

A = 0.04(0.01) + 0.06(0.01) = 0.001 m2

I =112

(0.04)(0.013) + (0.04)(0.01)(0.026 - 0.005)2

+112

(0.01)(0.063) + 0.01(0.06)(0.040 - 0.026)2 = 0.4773(10- 6) m4

s =PA

{ MxI

Assume tension failure,

180(106) =P

0.001+

0.226P(0.026)

0.4773(10- 6)

P = 13524 N = 13.5 kN

Assume compression failure,

-180(106) =P

0.001-

0.226P(0.070 - 0.026)0.4773(10- 6)

P = 9076 N = 9.08 kN (controls) Ans.

862.

Determine the maximum ram force P that can be applied to the clamp at D if the allowable normal stress for the material is sallow = 180 MPa.

D

a

40 mm

10 mm

60 mm 10 mm

200 mm

a P

Ans:Pmax = 9.08 kN

798

.

SolutionInternal Forces and Moments: As shown on FBD.

Fx = 0; 1.50 + Nx = 0 Nx = -15.0 kip

Fy = 0; Vy - 10.8 = 0 Vy = 10.8 kip

Fz = 0; Vz = 0

Mx = 0; Tx - 10.8(6) = 0 Tx = 64.8 kip # ft My = 0; My - 1.50(6) = 0 My = 9.00 kip # ft Mz = 0; 10.8(6) + Mz = 0 Mz = -64.8 kip # ftSection Properties:

A = p132 - 2.7522 = 1.4375p in2

Iy = Iz =p

4 134 - 2.7542 = 18.6992 in4

(QC)z = (QD)y = 0

(QC)y = (QD)z =4(3)

3p c 1

2 (p)1322 d - 4(2.75)

3p c 1

2(p)12.7522 d

= 4.13542 in3

J =p

2 134 - 2.7542 = 37.3984 in4

Normal Stress:

s =NA

-Mz y

Iz+

My z

Iy

sC =-1.50

1.4375p-

(-64.8)(12)(0)18.6992

+9.00(12)(2.75)

18.6992

= 15.6 ksi (T) Ans.

sD =-1.50

1.4375p-

(-64.8)(12)(3)18.6992

+9.00(12)(0)

18.6992

= 124 ksi (T) Ans.

863.

The uniform sign has a weight of 1500 lb and is supported by the pipe AB, which has an inner radius of 2.75 in. and an outer radius of 3.00 in. If the face of the sign is subjected to a uniform wind pressure of p = 150 lb>ft2, determine thestate of stress at points C and D. Show the results on adifferential volume element located at each of these points. Neglect the thickness of the sign, and assume that it is supported along the outside edge of the pipe.

3 ft

6 ft

12 ft

B

A

yx

z

CD

FE

150 lb/ft2

799

.

863. Continued

Shear Stress: The tranverse shear stress in the z and y directions and the torsional shear stress can be obtained using the shear formula and the torsion formula.

tV =VQ

It and ttwist =

Tr

J, respectively.

(txz)D = ttwist =64.8(12)(3)

37.3984= 62.4 ksi Ans.

(txy)D = tVy = 0 Ans.

(txy)C = tVy - ttwist

=10.8(4.13542)

18.6992(2)(0.25)-

64.8(12)(2.75)

37.3984

= -52.4 ksi Ans.

(txz)C = tVz = 0 Ans.

Ans:sC = 15.6 ksi (T), sD = 124 ksi (T),(txz)D = 62.4 ksi, (txy)D = 0,(txy)C = -52.4 ksi, (txz)C = 0

800

.

SolutionInternal Forces and Moments: As shown on FBD.

Fx = 0; 1.50 + Nx = 0 Nx = -1.50 kip

Fy = 0; Vy - 10.8 = 0 Vy = 10.8 kip

Fz = 0; Vz = 0

Mx = 0; Tx - 10.8(6) = 0 Tx = 64.8 kip # ft My = 0; My - 1.50(6) = 0 My = 9.00 kip # ft Mz = 0; 10.8(6) + Mz = 0 Mz = -64.8 kip # ftSection Properties:

A = p132 - 2.7522 = 1.4375p in2

Iy = Iz =p

4 134 - 2.7542 = 18.6992 in4

(QF)z = (QE)y = 0

(QF)y = (QE)z =4(3)

3p c 1

2 (p)1322 d - 4(2.75)

3p c 1

2 (p)12.7522 d

= 4.13542 in3

J =p

2 134 - 2.7542 = 37.3984 in4

Normal Stress:

s =NA

-Mzy

Iz+

My z

Iy

sF =-1.50

1.4375p-

(-64.8)(12)(0)18.6992

+9.00(12)(-3)

18.6992

= -17.7 ksi = 17.7 ksi (C) Ans.

sE =-1.50

1.4375p-

(-64.8)(12)(-3)18.6992

+9.00(12)(0)

18.6992

= -125 ksi = 125 ksi (C) Ans.

*864.

Solve Prob. 863 for points E and F.

3 ft

6 ft

12 ft

B

A

yx

z

CD

FE

150 lb/ft2

801

.

Shear Stress: The tranverse shear stress in the z and y directions and the torsional shear stress can be obtained using the shear formula and the torsion formula,

tV =VQ

It and ttwist =

Tr

J, respectively.

(txz)E = -ttwist = -64.8(12)(3)

37.3984= -62.4 ksi Ans.

(txy)E = tVy = 0 Ans.

(txy)F = tVy + ttwist

=10.8(4.13542)

18.6992(2)(0.25)+

64.8(12)(3)

37.3984

= 67.2 ksi Ans.

(txz)F = tVy = 0 Ans.

*864. Continued

Ans:sF = 17.7 ksi (C),sE = 125 ksi (C),(txz)E = -62.4 ksi,(txy)E = 0,(txy)F = 67.2 ksi,(txz)F = 0

802

.

Solution

I =14

(p)(0.014) = 7.85398(10-9) m4

QB = y-A= =4(0.01)

3pa1

2b(p)(0.012) = 0.66667(10-6) m3

QA = 0

sA =McI

=12(0.01)

7.85398(10-9)= 15.3 MPa Ans.

tA = 0 Ans.

sB = 0 Ans.

tB =VQB

It=

150(0.6667)(10-6)

7.85398(10-9)(0.02)= 0.637 MPa Ans.

865.

The pin support is made from a steel rod and has a diameter of 20 mm. Determine the stress components at points A and B and represent the results on a volume element located at each of these points.

BC

DA80 mm

150 N

Ans:sA = 15.3 MPa, tA = 0, sB = 0, tB = 0.637 MPa

803

.

Solution

I =14

(p)(0.014) = 7.85398(10-9) m4

QD = yA= =4(0.01)

3pa1

2b(p)(0.012) = 0.66667(10-6) m3

QC = 0

sC =McI

=12(0.01)

7.85398(10-9)= 15.3 MPa Ans.

tC = 0 Ans.

sD = 0 Ans.

tD =VQD

It=

150(0.6667)(10-6)

7.8539(10-9)(0.02)= 0.637 MPa Ans.

866.

Solve Prob. 865 for points C and D.

BC

DA80 mm

150 N

Ans:sC = 15.3 MPa, tC = 0, sD = 0, tD = 0.637 MPa

804

.

SolutionNormal stress due to axial force:

A = 2[0.5(3)] + 5(0.5) = 5.5 in2

sA =PA

=605.5

= 10.9090 psi (T)

Normal stress due to bending:

r- = 15 in. rA = 12 in. rB = 18 in.

LdAr

= b lnr2r1

= 3 ln12.512

+ 0.5 ln17.512.5

+ 3 ln18

17.5= 0.3752 in.

R =A

1dAr =5.5

0.3752= 14.6583 in.

r- - R = 0.3417 in.

(sA)b =M(R - rA)ArA(r

- - R)=

59.0(12)(14.6583 - 12)5.5(12)(0.3417)

= 83.4468 psi (T)

(sB)b =M(R - rB)ArB(r

- - R)=

59.0(12)(14.6583 - 18)5.5(18)(0.3417)

= -69.9342 psi = 69.9342 psi (C)

sA = 83.4468 + 10.9090 = 94.4 psi (T) Ans.

sB = 69.9342 - 10.9090 = 59.0 psi (C) Ans.

867.

The handle of the press is subjected to a force of 20 lb. Due to internal gearing, this causes the block to be subjected to a compressive force of 80 lb. Determine the normal-stress acting in the frame at points along the outside flanges A and B. Use the curved-beam formula to calculate the bending stress.

3 in.

0.8 ft

1.50 ft1 ft

A B

5 in.0.5 in. 0.5 in.

0.5 in.80 lb

20 lb

Ans:sA = 94.4 psi (T),sB = 59.0 psi (C)

805

.

*868.

The bar has a diameter of 40 mm. Determine the state of stress at point A and show the results on a differential volume element located at this point.


Fx = 0; Nx - 1200 = 0 Nx = 1200 N

Fy = 0; Vy - 800a35b = 0 Vy = 480 N

Fz = 0; Vz + 800a45b = 0 Vz = -640 N

Mx = 0; Tx = 0

My = 0; My + 800a45 b(0.2) = 0 My = -128 N# m

Mz = 0; Mz + 800a35b(0.2) = 0 Mz = -96 N # m


A = pc2 = p(0.022) = 0.4(10-3)p m2

Iy = Iz =p

4 c4 =

p

4(0.024) = 40.0(10-9)p m4

J =p

2c4 =

p

2(0.024) = 80.0(10-9)p m4

(QA)z = z- =A= =4(0.02)

3pcp

2(0.022) d = 5.3333(10-6) m3

(QA)y = 0

Normal Stress: For the combined loading, the normal stress at point A can be determined from

sx = sA =NxA

-MzyA

Iz+

MyzAIy

=1200

0.4(10-3)p-

(-96)(0.02)40.0(10-9)p

+(-128)(0)

40.0(10-9)p

= 16.23(106) Pa = 16.2 MPa (T) Ans.

Shear Stress: Since Tx = 0, the shear stress in the z and y directions is contributed by transverse shear stress only which can be obtained using the shear formula,

tV =VQ

It

(txz)A = (tV)z =-64035.3333(10-6)440.0(10-9)p(0.04)

= -0.6791(106) Pa = -0.679 MPa Ans.

(txy)A = (tV)y = 0 Ans.

Using these results, the state of stress at point A can be represented by the volume element shown in Fig. c.

200 mm

200 mm

y

z

x

B

A

800 N

34

5

1200 N

Ans:sA = 16.2 MPa (T),(txz)A = -0.679 MPa,(txy)A = 0

806

.

869.

Solve Prob. 868 for point B.

200 mm

200 mm

y

z

x

B

A

800 N

34

5

1200 N


Fx = 0; Nx - 1200 = 0 Nx = 1200 N

Fy = 0; Vy - 800a35b = 0 Vy = 480 N

Fz = 0; Vz + 800a45b = 0 Vz = -640 N

Mx = 0; Tx = 0

My = 0; My + 800a45 b(0.2) = 0 My = -128 N# m

Mz = 0; Mz + 800a35b(0.2) = 0 Mz = -96 N # m


A = pc2 = p(0.022) = 0.4(10-3)p m2

Iy = Iz =p

4 c4 =

p

4(0.024) = 40.0(10-9)p m4

J =p

2c4 =

p

2(0.024) = 80.0(10-9)p m4

(QB)y = y- =A= =4(0.02)

3pcp

2(0.022) d = 5.3333(10-6) m3

(QB)z = 0

Normal Stress: For the combined loading, the normal stress at point B can be determined from

sx = sB =NxA

-MzyB

Iz+

MyzBIy

=1200

0.4(10-3)p-

(-96)(0)40.0(10-9)p

+(-128)(0.02)40.0(10-9)p

= -19.42(106) Pa = 19.4 MPa (C) Ans.

Shear Stress: Since Tx = 0, the shear stress in z and y directions is contributed by

transverse shear stress only, which can be obtained using the shear formula, tV =VQ

It.

(txy)B = (tV)y =48035.3333(10-6)440.0(10-9)p(0.04)

= 0.5093(106) Pa = 0.509 MPa Ans.

(txz)B = (tV)z = 0 Ans.

Using these results, the state of stress at point B can be represented by the volume element shown in Fig. c.

Ans:sB = 19.4 MPa (C),(txy)B = 0.509 MPa,(txz)B = 0

807

.

Solution

A =p

4 (0.752) = 0.44179 in2

I =p

4 (0.3754) = 0.015531 in4

QA = 0

tA = 0 Ans.

sA =My c

I=

-1250(0.375)0.015531

= -30.2 ksi = 30.2 ksi (C) Ans.

870.

The 34-in.-diameter shaft is subjected to the loading shown. Determine the stress components at point A. Sketch the results on a volume element located at this point. The journal bearing at C can exert only force components Cy and Cz on the shaft, and the thrust bearing at D can exert force components Dx, Dy, and Dz on the shaft.

x

C

A

B

125 lb

2 in.

8 in.

8 in.125 lb

D

20 in.

20 in.

10 in.

z

y

2 in.

Ans:tA = 0, sA = 30.2 ksi (C)

808

.

Solution

A =p

4 (0.752) = 0.44179 in2

I =p

4 (0.3754) = 0.015531 in4

QB = yA =4(0.375)

3p a1

2b(p)(0.3752) = 0.035156 in3

sB = 0 Ans.

tB =VzQB

It=

125(0.035156)

0.015531(0.75)= 0.377 ksi Ans.

871.

Solve Prob. 870 for the stress components at point B.

x

C

A

B

125 lb

2 in.

8 in.

8 in.125 lb

D

20 in.

20 in.

10 in.

z

y

2 in.

Ans:sB = 0, tB = 0.377 ksi



809

.

SolutionThe location of the neutral surface from the center of curvature of the hook, Fig. a, can be determined from

R =A

LAdAr

where A = p(0.252) = 0.0625p in2

LAdAr

= 2p1r - 2r 2 - c22 = 2p11.75 - 21.752 - 0.2522 = 0.11278 in.Thus,

R =0.0625p0.11278

= 1.74103 in.

Then

e = r - R = 1.75 - 1.74103 = 0.0089746 in.

Referring to Fig. b, I and QA are

I =p

4 (0.254) = 0.9765625(10- 3)p in4

QA = 0

Consider the equilibrium of the FBD of the hooks cut segment, Fig. c.

d+ Fx = 0; N - 80 cos 45 = 0 N = 56.57 lb

+ cFy = 0; 80 sin 45 - V = 0 V = 56.57 lb

a+Mo = 0; M - 80 cos 45(1.74103) = 0 M = 98.49 lb # inThe normal stress developed is the combination of axial and bending stress. Thus,

s =NA

+M(R - r)

Ae r

Here, M = 98.49 lb # in since it tends to reduce the curvature of the hook. For point A, r = 1.5 in. Then

s =56.57

0.0625p+

(98.49)(1.74103 - 1.5)0.0625p(0.0089746)(1.5)

= 9.269(103) psi = 9.27 ksi (T) Ans.

The shear stress is contributed by the transverse shear stress only. Thus

t =VQA

It= 0 Ans.

The state of stress of point A can be represented by the element shown in Fig. d.

*872.

The hook is subjected to the force of 80 lb. Determine the state of stress at point A at section aa. The cross section is circular and has a diameter of 0.5 in. Use the curved-beam formula to calculate the bending stress.

a

a

80 lb

1.5 in.

A A

B

B

45

Ans:s = 9.27 ksi (T),t = 0

810

.

SolutionThe location of the neutral surface from the center of curvature of the hook, Fig. a, can be determined from

R =A

LA dAr

where A = p(0.252) = 0.0625p in2

LA dAr

= 2p 1r - 2r 2 - c22 = 2p 11.75 - 21.752 - 0.2522 = 0.11278 in.Thus,

R =0.0625p0.11278

= 1.74103 in

Then

e = r - R = 1.75 - 1.74103 = 0.0089746 in

Referring to Fig. b, I and QB are computed as

I =p

4 (0.254) = 0.9765625(10- 3)p in4

QB = yA =4(0.25)

3p cp

2 (0.252) d = 0.0104167 in3

Consider the equilibrium of the FBD of the hooks cut segment, Fig. c.

d+ Fx = 0; N - 80 cos 45 = 0 N = 56.57 lb

+ cFy = 0; 80 sin 45 - V = 0 V = 56.57 lb

a+Mo = 0; M - 80 cos 45 (1.74103) = 0 M = 98.49 lb # inThe normal stress developed is the combination of axial and bending stress. Thus,

s =NA

+M(R - r)

Ae r

Here, M = 98.49 lb # in since it tends to reduce the curvature of the hook. For point B, r = 1.75 in. Then

s =56.57

0.0625p+

(98.49)(1.74103 - 1.75)0.0625 p (0.0089746)(1.75)

= 1.48 psi (T) Ans.

The shear stress is contributed by the transverse shear stress only. Thus,

t =VQB

It=

56.57 (0.0104167)

0.9765625(10- 3)p (0.5)= 384 psi Ans.

The state of stress of point B can be represented by the element shown in Fig. d.

873.

The hook is subjected to the force of 80 lb. Determine the state of stress at point B at section aa. The cross section has a diameter of 0.5 in. Use the curved-beam formula to calculate the bending stress.

a

a

80 lb

1.5 in.

A A

B

B

45

Ans:s = 1.48 psi (T), t = 384 psi

811

.

Solution

L dAr = 2pa3.125 - 2(3.125)2 - (0.625)2 b = 0.395707

R =A

LdAr=

p(0.625)2

0.396707= 3.09343 in.

M = 800(3.09343) = 2.475(103)

s =M(R - r)Ar(r - R)

+PA

(st)max =2.475(103)(3.09343 - 2.5)

p(0.625)2(2.5)(3.125 - 3.09343)+

800

p(0.625)2= 15.8 ksi Ans.

(sc)max =2.475(103)(3.09343 - 3.75)

p(0.625)2(3.75)(3.125 - 3.09343)+

800

p(0.625)2= -10.5 ksi Ans.

R81.

The eye hook has the dimensions shown. If it supports a cable loading of 800 lb, determine the maximum normal stress at section aa and sketch the stress distribution acting over the cross section. Use the curved-beam formula to calculate the bending stress.

3.75 in. 2.5 in.

a

800 lb

1.25 in.

a

800 lb

Ans:(st)max = 15.8 ksi, (sc)max = -10.5 ksi

812

.

SolutionSupport Reactions: Referring to the free-body diagram of member BC shown in Fig. a,a+MB = 0; F sin 45(1) - 20(9.81)(2) = 0 F = 554.94 N

S+ Fx = 0; 554.94 cos 45 - Bx = 0 Bx = 392.4 N

+ cFy = 0; 554.94 sin 45 - 20(9.81) - By = 0 By = 196.2 N

Internal Loadings: Consider the equilibrium of the free-body diagram of the right segment shown in Fig. b.

S+ Fx = 0; N - 392.4 = 0 N = 392.4 N

+ cFy = 0; V - 196.2 = 0 V = 196.2 Na+MC = 0; 196.2(0.5) - M = 0 M = 98.1 N # mSection Properties: The cross-sectional area and the moment of inertia of the cross section are

A = 0.05(0.075) = 3.75110- 32 m2

I =112

(0.05)10.07532 = 1.7578110- 62 m4

Referring to Fig. c, QE is

QE = yA = 0.025(0.025)(0.05) = 31.25110- 62 m3Normal Stress: The normal stress is the combination of axial and bending stress. Thus,

s =NA

{My

I

For point E, y = 0.0375 - 0.025 = 0.0125 m. Then

sE =392.4

3.75110- 32 +98.1(0.0125)

1.7578110- 62 = 802 kPa Ans.Shear Stress: The shear stress is contributed by transverse shear stress only. Thus,

tE =VQA

It=

196.2331.25110- 62 41.7578110- 62(0.05) = 69.8 kPa Ans.

The state of stress at point E is represented on the element shown in Fig. d.

R82.

The 20-kg drum is suspended from the hook mounted on the wooden frame. Determine the state of stress at point E on the cross section of the frame at sectionaa. Indicate the results on an element.

1 m

1 m

1 m

b

a

a

b

CB

A

30

1 m0.5 m0.5 m

50 mm

75 mm25 mm

Section a a

E

75 mm

75 mm

25 mmSection b b

FD

Ans:sE = 802 kPa, tE = 69.8 kPa

813

.

R83.

The 20-kg drum is suspended from the hook mounted on the wooden frame. Determine the state of stress at point F on the cross section of the frame at section bb. Indicate the results on an element.

1 m

1 m

1 m

b

a

a

b

CB

A

30

1 m0.5 m0.5 m

50 mm

75 mm25 mm

Section a a

E

75 mm

75 mm

25 mmSection b b

FD

SolutionSupport Reactions: Referring to the free-body diagram of the entire frame shown in Fig. a,a+MA = 0; FBD sin 30(3) - 20(9.81)(2) = 0 FBD = 261.6 N

+ cFy = 0; Ay - 261.6 cos 30 - 20(9.81) = 0 Ay = 422.75 N

S+ Fx = 0; Ax - 261.6 sin 30 = 0 Ax = 130.8 N

Internal Loadings: Consider the equilibrium of the free-body diagram of the lower cut segment, Fig. b.

S+ Fx = 0; 130.8 - V = 0 V = 130.8 N

+ cFy = 0; 422.75 - N = 0 N = 422.75 N

a+MC = 0; 130.8(1) - M = 0 M = 130.8 N # mSection Properties: The cross-sectional area and the moment of inertia about the centroidal axis of the cross section are

A = 0.075(0.075) = 5.625110- 32 m2

I =112

(0.075)10.07532 = 2.6367110- 62 m4

Referring to Fig. c, QE is

QF = yA = 0.025(0.025)(0.075) = 46.875110- 62 m3Normal Stress: The normal stress is the combination of axial and bending stress. Thus,

s =NA

{My

I

For point F, y = 0.0375 - 0.025 = 0.0125 m. Then

sF =-422.75

5.625110- 32 -130.8(0.0125)

2.6367110- 62= -695.24 kPa = 695 kPa (C) Ans.

Shear Stress: The shear stress is contributed by transverse shear stress only. Thus,

tA =VQA

It=

130.8 c 46.875110- 62 d2.6367110- 62(0.075) = 31.0 kPa Ans.

The state of stress at point A is represent

Solution manual for mechanics of materials 10th edition hibbeler sample

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