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 2.13, Problem 107P

(a)

To determine

Find the maximum deflection at point C of the cylindrical portions.

(a)

Expert Solution
Check Mark

Answer to Problem 107P

The maximum deflection at point C of the cylindrical portions is 0.292mm_.

Explanation of Solution

Given information:

The cross sectional area AC and BC of each portions is 1,750mm2.

The modulus of elasticity (E) for portion AC is 200GPa

The yield stress (σy) for portion AC is 250MPa

The modulus of elasticity (E) for portion CB is 200GPa

The yield stress (σy) for portion CB is 345MPa

Calculation:

Calculate the displacement at point C to cause yielding of AC using the relation:

δC,Y=LACεY,AC=LACσY,ACE (1)

Here, LAC is length of portion AC and εY,AC is strain yielding of AC.

Substitute 190mm for LAC and 250MPa for σy, and 200GPa for E in Equation (1).

δC,Y=190mm(1m103mm)×250MPa(106Pa1MPa)200GPa(109Pa1GPa)=0.190×250×106200×109=0.2375×103m

Find the corresponding force along AC using the relation as follows:

FAC=AσY,AC (2)

Substitute 1,750mm2 for A and 250MPa for σY,AC in Equation (2).

FAC=1,750mm2(1m103mm)2×250MPa(106Pa1MPa)=1750×106×250×106=437.5×103N

Find the corresponding force along CB using the relation as follows:

FCB=EAδcLCB (3)

Substitute 1,750mm2 for A, 190mm for LCB , 0.2375×103m for δc, 200GPa for E in Equation (3).

FCB=200GPa(109Pa1GPa)×1,750mm2(1m103mm)2×0.2375×103m190mm(1m103mm)=200×109×1750×106×0.2375×1030.190=437.5×103N

Sketch the element C as shown in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 2.13, Problem 107P

Refer to Figure 1.

Find the value of P using equilibrium element.

FAC(FCB+PY)=0PY=FACFCB (4)

Substitute 437.5×103N for FCB and 437.5×103N for FAC in Equation (4).

PY=437.5×103(437.5×103)=875×103N

Since the applied load,

P=975×103N>875×103N Portion AC yields.

Refer to Figure 1.

Find the force along CB as follows:

FCB=FACP

Substitute 437.5×103N for FAC and 975×103N for P.

FCB=437.5×103975×103N=537.5×103N

Determine the deflection at point C using the relation:

δC=FCBLCDEA (5)

Substitute 537.5×103N for FCB, 190mm for LCD, 200GPa for E, and 1,750mm2 for A in Equation (5).

δC=537.5×103×190mm(1m103mm)200GPa(109Pa1GPa)×1750mm2(1m103mm)2=537.5×103×0.190200×109×1750×106=0.29179×103m(103mm1m)=0.292mm

Thus, the maximum deflection at point C of the cylindrical portions is 0.292mm_.

(b)

To determine

Find the maximum stress for each portion of rod.

(b)

Expert Solution
Check Mark

Answer to Problem 107P

The maximum stress of rod AC is 250MPa_.

The maximum stress of rod BC is 307MPa_.

Explanation of Solution

Calculation:

Refer part a.

The maximum stress of rod AC is 250MPa

Therefore, the maximum stress of rod AC is 250MPa_.

Determine the maximum stress at point BC using the relation:

σBC=FBCA (6)

Substitute 537.5×103N for FCB and 1,750mm2 for A in Equation (6).

σBC=537.5×1031750mm2(1m103mm)2=307.14×106Pa(1MPa106Pa)=307MPa

Thus, the maximum stress of rod BC is 307MPa_.

(c)

To determine

Find the permanent deflection at point C.

(c)

Expert Solution
Check Mark

Answer to Problem 107P

The permanent deflection at point C is 0.0272mm_.

Explanation of Solution

Write the expression of deflection and force for unloading as follows:

δ=PACLACEA=PCBLCBEA

PCB=PACLACLCB=PAC

The value of P is 975×103N.

P=PACPCB

Substitute PAC for PCB. And 975×103N for P.

975×103=PAC(PAC)975×103=2PACPAC=487.5×103N

Determine the deflection using the relation.

δ=PACLACEA (7)

Substitute 1,750mm2 for A, 190 mm for LAC, 200GPa for E, and 487.5×103N for PAC in Equation (7).

δ=487.5×103×190mm(1m103mm)200GPa(109pa1GPa)(1750mm2×(1m103mm)2)=487.5×103×0.190200×109(1750×106)=0.26464×103m

Find the permanent deflection using the relation:

δP=δmδ (8)

Substitute 0.26464×103m for δ and 0.29179×103 for δm in Equation (8).

δP=0.26464×1030.26464×103=0.02715×103m(103mm1m)=0.0272mm

Thus, the permanent deflection at point C is 0.0272mm_.

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Chapter 2 Solutions

Mechanics of Materials, 7th Edition

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