Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
10th Edition
ISBN: 9780073398204
Author: Richard G Budynas, Keith J Nisbett
Publisher: McGraw-Hill Education
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Concept explainers

Question
Chapter 3, Problem 73P

(a)

To determine

The force FB, assuming the shaft is running at constant speed.

(a)

Expert Solution
Check Mark

Answer to Problem 73P

The force FB, assuming the shaft is running at constant speed is 22810N.

Explanation of Solution

The figure below shows the free body diagram of pulley A.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 3, Problem 73P , additional homework tip  1

Figure-(1)

The figure below shows the free body diagram of pulley B.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 3, Problem 73P , additional homework tip  2

Figure-(2)

Convert the forces from kN to N.

FA=11kN=(11kN)(1000N1kN)=11000N

Calculate the force FB, using the net torque equation.

T=0(FA×dA2×cosθ1)+(FB×dB2×cosθ2)=0FB=FA(dAcosθ1dBcosθ2) (I)

Here, the force acting on pulley A is FA, the diameter of pulley A is dA, the angle at which force acts on pulley A is θ1, the force acting on pulley B is FB, the diameter of pulley B is dB and the angle at which force acts on pulley B is θ2.

Conclusion:

Substitute 11000N for FA, 600mm for dA, 20° for θ1, 300mm for dB and 25° for θ2 in Equation (I).

FB=(11000N)×(600mm(cos20°)300mm(cos25°))=(11000N)(2.0736)=22810.39N22810N

Thus, the force FB, assuming the shaft is running at constant speed is 22810N.

(b)

To determine

The bearing reaction forces, assuming the bearing act as simple supports.

(b)

Expert Solution
Check Mark

Answer to Problem 73P

The bearing reaction forces, assuming the bearing act as simple supports at O is 7926N and at C is 13657N.

Explanation of Solution

Write the expression for moment about bearing O in z- direction.

MOz=0[(FAsinθ1×lOA)(FBsinθ2×(lOA+lAB))+RCy×(lOA+lAB+lBC)]=0RCy=[(FBsinθ2×(lOA+lAB))+(FAsinθ1×lOA)](lOA+lAB+lBC) (II)

Here, the reaction force at bearing C in y- direction is RCy, the distance between O and A is lOA, the distance between A and B is lAB and the distance between B and C is lBC.

Write expression for the equation to balance the forces in y- direction.

Fy=0ROyFAsinθ1+RCyFBsinθ2=0ROy=FBsinθ2RCy+FAsinθ1 (III)

Here, the reaction force at bearing O in y- direction is ROy.

Write expression for the moment about bearing O in y- direction.

MOy=0[(FAcosθ1×lOA)(FBcosθ2×(lOA+lAB))RCz×(lOA+lAB+lBC)]=0RCz=[(FAcosθ1×lOA)(FBcosθ2×(lOA+lAB))](lOA+lAB+lBC) (IV)

Here, the reaction force at bearing C in z- direction is RCz.

Write expression for the equation to balance the forces in z- direction.

Fz=0ROzFAcosθ1+RCz+FBcosθ2=0ROz=FAsinθ1RCzFBcosθ2 (V)

Here, the reaction force at bearing O in z- direction is ROz.

Calculate the reaction forces at bearing O.

RO=ROy2+ROz2 (VI)

Here, the reaction force at bearing O is RO.

Calculate the reaction forces at bearing C.

RC=RCy2+RCz2 (VII)

Here, the reaction force at bearing C is RC.

Conclusion:

Substitute 22810N for FB, 20° for θ1, 11000N for FA, 25° for θ2, 400mm for lOA, 350mm for lAB and 300mm for lBC in Equation (II).

RCy=[((22810N)(sin25°)(400mm+350mm))+((11000N)(sin20°)(400mm))](400mm+350mm+300mm)=8734830.543Nmm1050mm=(8318.8862N)(1kN1000N)8.319kN

Convert RCy from kN to N.

RCy=8.319kN=(8.319kN)(1000N1kN)=8319N

Substitute 22810N for FB, 25° for θ2, 8319N for RCy, 11000N for FA and 20° for θ1 in Equation (III).

ROy=[(22810N)(sin25°)(8319N)+(11000N)(sin20°)]=9639.9225N8319N+3762.2215N=(5083.144N)5083N

Substitute 22810N for FB, 20° for θ1, 11000N for FA, 25° for θ2, 400mm for lOA, 350mm for lAB and 300mm for lBC in Equation (IV).

RCz=[((11000N)(cos20°)(400mm))((22810N)(cos25°)(400mm+350mm))](400mm+350mm+300mm)=11370012.94Nmm1050mm10830N

Substitute 11000N for FA, 20° for θ1, 10830N for RCz, 22810N for FB and 25° for θ2 in Equation (V).

ROz=(11000N)(sin20°)+10830N(22810N)(cos25°)=14592.22158N20672.88062N=6080.66N6081N

Substitute 5083N for ROy and 6081N for ROz in Equation (VI).

RO=(5083N)2+(6081N)2=25836889N2+36978561N2=62815450N27926N

Substitute 8319N for RCy and 10830N for RCz in Equation (VII).

RC=(8319N)2+(10830N)2=186494661N2=13656.30481N13657N

Thus, the bearing reaction forces, assuming the bearing act as simple supports at O is 7926N and at C is 13657N.

(c)

To determine

The shear force and bending moment diagram for the shaft.

(c)

Expert Solution
Check Mark

Answer to Problem 73P

The shear force and bending moment diagram for the shaft in y- direction is as follows.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 3, Problem 73P , additional homework tip  3

Figure-(2)

The shear force and bending moment diagram for the shaft in z- direction is as follows.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 3, Problem 73P , additional homework tip  4

Figure-(3)

Explanation of Solution

The calculations for shear force and bending moment diagram in y- direction.

Calculate the shear force at O in y- direction.

SFOy=ROy (VIII)

Here, the shear force at O in y- direction is SFOy.

Calculate the shear force at A in y- direction.

SFAy=SFOyFAsinθ1 (IX)

Here, the shear force at A in y- direction is SFAy.

Calculate the shear force at B in y- direction.

SFBy=SFAy(FBsinθ2) (X)

Here, the shear force at B in y- direction is SFBy.

Calculate the shear force at C in y- direction.

SFCy=SFBy+RCy (XI)

Here, the shear force at B in y- direction is SFBy.

Calculate the moment at O and C.

MO=MC=0 (XII)

Here, the moment at O is MO and the moment at C is MC.

Calculate the moment at A in y- direction.

MAy=SFOy×lOA (XIII)

Here, the moment at A in y- direction is MAy.

Calculate the moment at B in y- direction.

MBy=FBsinθ2×lCB (XIV)

Here, the moment at B in y- direction is MBy.

The calculations for shear force and bending moment diagram in z- direction.

Calculate the shear force at O in y- direction.

SFOz=ROz (XV)

Here, the shear force at O in z- direction is SFOz.

Calculate the shear force at A in z- direction.

SFAz=SFOz+FAcosθ1 (XVI)

Here, the shear force at A in z- direction is SFAz.

Calculate the shear force at B in z- direction.

SFBz=SFAzFBcosθ2 (XVII)

Here, the shear force at B in z- direction is SFBz.

Calculate the shear force at C in z- direction.

SFCz=SFBzRCz (XVIII)

Here, the shear force at C in z- direction is SFCz.

Calculate the moment at O and C.

MO=MC=0 (XIX)

Here, the moment at O is MO and the moment at C is MC.

Calculate the moment at A in z- direction.

MAz=SFAz×lOA (XX)

Here, the moment at A in z- direction is MAz.

Calculate the moment at B in z- direction.

MBz=RCz×lBC (XXI)

Here, the moment at B in z- direction is MBz.

Conclusion:

Substitute 5083N for ROy in Equation (VIII).

SFOy=5083N

Substitute 5083N for SFOy, 11000N for FA and 20° for θ1 in Equation (IX.)

SFAy=5083N(11000N)(sin20)=1320.778N=1321N

Substitute 1321N for SFAy, 22810N for FB and 25° for θ2 in Equation (X).

SFBy=(1321N)((22810N)(sin25°))=8318.92255N8319N

Substitute 8319N for SFBy and 8319N for RCy in Equation (XI).

SFCy=8319N+8319N=0N

Substitute 5083N for SFOy and 400mm for lOA in Equation (XIII).

MAy=(5083N)(400mm)=(5083N)(400mm)(1m1000mm)=2033.2Nm2033Nm

Substitute 22810N for FB, 25° for θ2 and 300mm for lBC in Equation (XIV).

MBy=(22810N)(sin25°)(300mm)=(22810N)(sin25°)(300mm)(1m1000mm)2892Nm

Thus, the shear force and bending moment diagram in y- direction is as follow.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 3, Problem 73P , additional homework tip  5

Figure-(4)

Substitute 6081N for ROz in Equation (XV).

SFOz=6081N

Substitute 6081N for SFOz, 11000N for FA and 20° for θ1 in Equation (XVI).

SFAz=6081N+(11000N)(cos20°)=4255.6184256N

Substitute 4256N for SFAz, 22810N for FB and 25° for θ2 in Equation (XVII).

SFBz=4256N(22810N)(cos25°)=16416.88N16417N

Substitute 16417N for SFBz and 10830N for RCz in Equation (XVIII).

SFCz=(16417N)(10830N)=16417N+10830N=5587N

Substitute 4256N for SFAz and 400mm for lOA in Equation (XX).

MAz=(4256N)(400mm)=(4256N)(400mm)(1m1000mm)=1702.4Nm1702Nm

Substitute 10830N for FB and 300mm for lBC in Equation (XXI).

MBz=(10830N)(300mm)=(10830N)(300mm)(1m1000mm)=3249Nm

Thus, the shear force and bending moment diagram in z- direction is as follow.

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering), Chapter 3, Problem 73P , additional homework tip  6

Figure-(5)

(d)

To determine

The bending stress at point of maximum bending moment.

The shear stress at point of maximum bending moment.

(d)

Expert Solution
Check Mark

Answer to Problem 73P

The bending stress at point of maximum bending moment is 355MPa.

The shear stress at point of maximum bending moment is 127MPa.

Explanation of Solution

Write the net moment at A.

MA=MAy2+MAz2 (XXII)

Here, the net moment at A is MA.

Write the net moment at B.

MB=MBy2+MBz2 (XXIII)

Here, the net moment at C is MC.

Write the torque transmitted by shaft from A to B.

T=FAcosθ1×dA2 . (XXIV)

Here, the torque transmitted by shaft from A to B is T.

Calculate the bending stress.

σ=32MBπd3 (XXV)

Here, the bending stress is σ and diameter of shaft is d.

Calculate the shear stress.

τ=16Tπd3 (XVI)

Here, the shear stress is τ.

Conclusion:

Substitute 2033Nm for MAy and 1702Nm for MAz in Equation (XXII).

MA=(2033Nm)2+(1702Nm)2=(7029893)Nm2652Nm

Substitute 2892Nm for MBy and 3249Nm for MBz in Equation (XXIII).

MB=(2892Nm)2+(3249Nm)2=18919665N2m2=4349.674Nm4350Nm

Since, MB>MA, the critical location is at B.

Substitute 11000N for FA, 20° for θ1 and 600mm for dA in Equation (XXIV).

T=(11000N)(cos20°)(600mm)2=(11000N)(cos20°)(600mm)(1m1000mm)2=3100.9856Nm3101Nm

Substitute 4350Nm for MB and 50mm for d in Equation (XXV).

σ=(32)(4350Nm)π(50mm)3=(32)(4350Nm)π((50mm)(1m1000mm))3=139200Nm(3.92699×104)m3(1MPa106N/m2)355MPa

Thus, the bending stress at point of maximum bending moment is 355MPa.

Substitute 3101Nm for T and 50mm for d in Equation (XXVI).

τ=16×3101Nmπ(50mm)3=(16)(3101Nm)π((50mm)(1m1000mm))3=49616Nm(3.92699×104)m3(1MPa106N/m2)127MPa

Thus, the shear stress at point of maximum bending moment is 127MPa.

(e)

To determine

The principal stresses at point of maximum bending moment.

The maximum shear stress at point of maximum bending moment.

(e)

Expert Solution
Check Mark

Answer to Problem 73P

The principal stresses at point of maximum bending moment are 396MPa and 41MPa.

The maximum shear stress at point of maximum bending moment is 218MPa.

Explanation of Solution

Calculate the maximum principal stress.

σ1=σ2+(σ2)2+τ2 (XXVII)

Here, the maximum principal stress is σ1.

Calculate the minimum principal stress.

σ2=σ2(σ2)2+τ2 (XXVIII)

Here, the minimum principal stress is σ2.

Calculate the maximum shear stress.

τmax=(σ2)2+τ2 (XXIX)

Here, maximum shear stress is τmax.

Conclusion:

Substitute 355MPa for σ and 127MPa for τ in Equation (XXVII).

σ1=355MPa2+(355MPa2)2+(127MPa)2=177.5MPa+218.255MPa396MPa

Substitute 355MPa for σ and 127MPa for τ n Equation (XXVIII).

σ2=355MPa2(355MPa2)2+(127MPa)2=177.5MPa218.255MPa41MPa

Thus, the principal stresses at point of maximum bending moment are 396MPa and 41MPa.

Substitute 355MPa for σ and 127MPa for τ in Equation (XXIX).

τmax=(355MPa2)2+(127MPa)2=47635.25MPa2=218.255MPa218MPa

Thus, the maximum shear stress at point of maximum bending moment is 218MPa.

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Chapter 3 Solutions

Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)

Ch. 3 - Repeat Prob. 37 using singularity functions...Ch. 3 - Repeat Prob. 38 using singularity functions...Ch. 3 - For a beam from Table A9, as specified by your...Ch. 3 - A beam carrying a uniform load is simply supported...Ch. 3 - For each of the plane stress states listed below,...Ch. 3 - Repeat Prob. 315 for: (a)x = 28 MPa, y = 7 MPa, xy...Ch. 3 - Repeat Prob. 315 for: a) x = 12 kpsi, y = 6 kpsi,...Ch. 3 - For each of the stress states listed below, find...Ch. 3 - Repeat Prob. 318 for: (a)x = 10 kpsi, y = 4 kpsi...Ch. 3 - The state of stress at a point is x = 6, y = 18, z...Ch. 3 - The state of stress at a point is x = 6, y = 18, z...Ch. 3 - Repeat Prob. 320 with x = 10, y = 40, z = 40, xy =...Ch. 3 - A 34-in-diameter steel tension rod is 5 ft long...Ch. 3 - Repeat Prob. 323 except change the rod to aluminum...Ch. 3 - A 30-mm-diameter copper rod is 1 m long with a...Ch. 3 - A diagonal aluminum alloy tension rod of diameter...Ch. 3 - Repeat Prob. 326 with d = 16 mm, l = 3 m, and...Ch. 3 - Repeat Prob. 326 with d = 58 in, l = 10 ft, and...Ch. 3 - Electrical strain gauges were applied to a notched...Ch. 3 - Repeat Prob. 329 for a material of aluminum. 3-29...Ch. 3 - The Roman method for addressing uncertainty in...Ch. 3 - Using our experience with concentrated loading on...Ch. 3 - The Chicago North Shore Milwaukee Railroad was an...Ch. 3 - For each section illustrated, find the second...Ch. 3 - 3-35 to 3-38 For the beam illustrated in the...Ch. 3 - 3-35 to 3-38 For the beam illustrated in the...Ch. 3 - 3-35 to 3-38 For the beam illustrated in the...Ch. 3 - 3-35 to 3-38 For the beam illustrated in the...Ch. 3 - The figure illustrates a number of beam sections....Ch. 3 - A pin in a knuckle joint canning a tensile load F...Ch. 3 - Repeat Prob. 3-40 for a = 6 mm, b = 18 mm. d = 12...Ch. 3 - For the knuckle joint described in Prob. 3-40,...Ch. 3 - The figure illustrates a pin tightly fitted into a...Ch. 3 - For the beam shown, determine (a) the maximum...Ch. 3 - A cantilever beam with a 1-in-diameter round cross...Ch. 3 - Consider a simply supported beam of rectangular...Ch. 3 - In Prob. 346, h 0 as x 0, which cannot occur. 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