Mechanics of Materials, 7th Edition
Mechanics of Materials, 7th Edition
7th Edition
ISBN: 9780073398235
Author: Ferdinand P. Beer, E. Russell Johnston Jr., John T. DeWolf, David F. Mazurek
Publisher: McGraw-Hill Education
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Chapter 4.9, Problem 149P

(a)

To determine

Find the value of θ for which the stress at D reaches its largest value.

(a)

Expert Solution
Check Mark

Answer to Problem 149P

The value of θ for which the stress at D reaches its largest value is 53.1°_.

Explanation of Solution

Given information:

The radius of the circular plate is R=125mm.

The width of the rectangular post is b=200mm and the height of the rectangular post is d=150mm.

The applied force is P=4kN.

Calculation:

Sketch the cross section of disk as shown in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 4.9, Problem 149P , additional homework tip  1

Refer to Figure 1.

Calculate the moment along x direction (Mx) as shown below.

Mx=PRsinθ

Substitute 4kN for P and 125mm for R.

Mx=4kN×1,000N1kN×125mm×1m1,000mm×sinθ=500sinθ

Calculate the moment along z direction (Mz) as shown below.

Mz=PRcosθ

Substitute 4kN for P and 125mm for R.

Mz=4kN×1,000N1kN×125mm×1m1,000mm×cosθ=500cosθ

Sketch the cross section of the rectangular post as shown in Figure 2.

Mechanics of Materials, 7th Edition, Chapter 4.9, Problem 149P , additional homework tip  2

Refer to Figure 2.

Calculate the area (A) of the rectangular cross section as shown below.

A=bd

Substitute 200mm for b and 150mm for d.

A=200×150=30×103mm2×(1m1,000mm)2=30×103m2

Calculate the moment of inertia along x direction (Ix) as shown below.

Ix=bd312

Substitute 200mm for b and 150mm for d.

Ix=112×200×1503=56.25×106mm4×(1m1,000mm)4=56.25×106m4

Calculate the moment of inertia along z direction (Iz) as shown below.

Iz=db312

Substitute 200mm for b and 150mm for d.

Iz=112×150×2003=100×106mm4×(1m1,000mm)4=100×106m4

The location of point D along x direction is xD=100mm.

The location of point D along z direction is zD=75mm.

Calculate the stress (σ) as shown below.

σ=PAMxzIx+MzxIz (1)

Substitute PRsinθ for Mx and PRcosθ for Mz.

σ=PA+PRsinθ×zIxPRcosθ×xIz=P(1A+RzsinθIxRxcosθIz)

The stress σ to be maximum, when dσdθ=0 with z=zD and x=xD.

Calculate the stress at point D (σD) as shown below.

Substitute zD for z and xD for x.

σD=P(1A+RzDsinθIxRxDcosθIz)

Calculate the value of θ as shown below.

Differentiate both sides of the Equation with respect to θ.

dσDdθ=P(0+RzDIx(cosθ)RxDIz(sinθ))=P(RzDcosθIx+RxDsinθIz)

Substitute 0 for dσDdθ.

P(RzDcosθIx+RxDsinθIz)=0RzDcosθIx=RxDsinθIzsinθcosθ=IzzDIxxDtanθ=IzzDIxxD

Substitute 100×106mm4 for Iz, 56.25×106mm4 for Ix, 75mm for zD, and 100mm for xD.

tanθ=100×106×(75)56.25×106×100=1.3333θ=tan1(1.3333)=53.1°

Therefore, the value of θ for which the stress at D reaches its largest value is 53.1°_.

(b)

To determine

Find the values of stress at A, B, C, and D.

(b)

Expert Solution
Check Mark

Answer to Problem 149P

The stress at A is σA=700kPa_.

The stress at B is σB=100kPa_.

The stress at C is σC=133.3kPa_.

The stress at D is σD=967kPa_.

Explanation of Solution

Given information:

The radius of the circular plate is R=125mm.

The width of the rectangular post is b=200mm and the height of the rectangular post is d=150mm.

The applied force is P=4kN.

Calculation:

Refer to part (a).

The value of θ for which the stress at D reaches its largest value is 53.1°.

The bending moment along x direction is Mx=500sinθ.

Substitute 53.1° for θ.

Mx=500sin53.1°=400Nm

The bending moment along z direction is Mz=500cosθ.

Substitute 53.1° for θ.

Mz=500cos53.1°=300Nm

Calculate the stress at A (σA) as shown below.

Refer to Figure 2 in part (a).

The location of point A along x direction is xA=100mm.

The location of point A along z direction is zA=75mm.

Substitute 4kN for P, 30×103mm2 for A, 56.25×106mm4 for Ix, 100×106mm4 for Iz, 100mm for xA, 75mm for zA, 400Nm for Mx, and 300Nm for Mz in Equation (1).

σA=(4kN×1,000N1kN30×103mm2(400Nm×1,000mm1m)×75mm56.25×106mm4+(300Nm×1,000mm1m)(100mm)100×106mm2)=0.1333+0.5333+0.3=0.7N/mm2×1,000kPa1N/mm2=700kPa

Hence, the stress at A is σA=700kPa_.

Calculate the stress at B (σB) as shown below.

Refer to Figure 2 in part (a).

The location of point B along x direction is xB=100mm.

The location of point B along z direction is zB=75mm.

Substitute 4kN for P, 30×103mm2 for A, 56.25×106mm4 for Ix, 100×106mm4 for Iz, 100mm for xB, 75mm for zB, 400Nm for Mx, and 300Nm for Mz in Equation (1).

σB=(4kN×1,000N1kN30×103mm2(400Nm×1,000mm1m)×75mm56.25×106mm4+(300Nm×1,000mm1m)×100mm100×106mm2)=0.1333+0.53330.3=0.1N/mm2×1,000kPa1N/mm2=100kPa

Hence, the stress at B is σB=100kPa_.

Calculate the stress at C (σC) as shown below.

Refer to Figure 2 in part (a).

The location of point C along x direction is xC=0.

The location of point C along z direction is zC=0.

Substitute 4kN for P, 30×103mm2 for A, 56.25×106mm4 for Ix, 100×106mm4 for Iz, 0 for xC, 0 for zC, 400Nm for Mx, and 300Nm for Mz in Equation (1).

σC=(4kN×1,000N1kN30×103mm2(400Nm×1,000mm1m)×056.25×106mm4+(300Nm×1,000mm1m)×0100×106mm2)=0.1333+0+=0.1333N/mm2×1,000kPa1N/mm2=133.3kPa

Hence, the stress at C is σC=133.3kPa_.

Calculate the stress at D (σD) as shown below.

Refer to Figure 2 in part (a).

The location of point D along x direction is xD=100mm.

The location of point D along z direction is zD=75mm.

Substitute 4kN for P, 30×103mm2 for A, 56.25×106mm4 for Ix, 100×106mm4 for Iz, 100mm for xD, 75mm for zD, 400Nm for Mx, and 300Nm for Mz in Equation (1).

σD=(4kN×1,000N1kN30×103mm2(400Nm×1,000mm1m)×(75mm)56.25×106mm4+(300Nm×1,000mm1m)×100mm100×106mm2)=0.13330.53330.3=0.9666N/mm2×1,000kPa1N/mm2=967kPa

Therefore, the stress at D is σD=967kPa_.

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Mechanics of Materials, 7th Edition

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