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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Textbook Question
Chapter 11, Problem 133RP

A uniform rod of flexural rigidity EI is bent and loaded as shown. Determine (a) the horizontal deflection of point D, (b) the slope at point D.

Chapter 11, Problem 133RP, A uniform rod of flexural rigidity EI is bent and loaded as shown. Determine (a) the horizontal

Fig. P11.133

(a)

Expert Solution
Check Mark
To determine

Find the horizontal deflection of point D.

Answer to Problem 133RP

The horizontal deflection of point D is δD=5PL33EI_.

Explanation of Solution

Given information:

The flexural rigidity of the rod is EI.

Calculation:

Apply a couple MD at D.

Sketch the Free Body Diagram as shown in Figure 1.

Mechanics of Materials, 7th Edition, Chapter 11, Problem 133RP

Refer to Figure 1.

Apply the Equations of Equilibrium as shown below.

Summation of forces along vertical direction is Equal to zero.

Fy=0RAy=0

Summation of forces along horizontal direction is Equal to zero.

Fx=0RAx+P=0RAx=P

Take moment about A is Equal to zero.

MA=0MAMD=0MA=MD

Calculate the strain energy (U) as shown below.

U=UAB+UBC+UCD=0lMAB22EIdx+0lMBC22EIdx+0lMCD22EIdv (1)

Calculate the deflection of point D (δD) as shown below.

Differentiate both side of the Equation (1) with respect to P as shown below.

δD=UP=12EI(0l2MABMABPdx+0l2MBCMBCPdx+0L22MCDMCDPdv)=1EI(0lMABMABPdx+0lMBCMBCPdx+0lMCDMCDPdv) (2)

Consider the member AB.

The bending moment over portion AB is MAB=MD+Py (3)

Differentiate both side of the Equation (3) with respect to P as shown below.

MABP=y

Differentiate both side of the Equation (3) with respect to MD as shown below.

MABMD=1

Consider the member BC.

The bending moment over portion BC is MBC=MD+Pl (4)

Differentiate both side of the Equation (4) with respect to P as shown below.

MBCP=l

Differentiate both side of the Equation (4) with respect to MD as shown below.

MBCMD=1

Consider the member CD.

The bending moment over portion CD is MCD=MD+Py (5)

Differentiate both side of the Equation (5) with respect to P as shown below.

MCDP=y

Differentiate both side of the Equation (5) with respect to MD as shown below.

MCDMD=1

Calculate the deflection of point D (δD) as shown below.

Substitute MD+Py for MAB, y for MABP, MD+Pl for MBC, l for MBCMD, MD+Py for MCD, and y for MCDMD and apply the limits in Equation (2).

δD=1EI(0l(MD+Py)ydy+0l(MD+Pl)ldx+0l(MD+Py)ydy)

Substitute 0 for MD.

δD=1EI(0lPy2dy+0lPl2dx+0lPy2dy)=PEI((y33)0l+(l2x)0l+(y33)0l)=Pl33EI(1+3+1)=5Pl33EI

Therefore, the horizontal deflection of point D is δD=5PL33EI_.

(b)

Expert Solution
Check Mark
To determine

Find the slope at point D.

Answer to Problem 133RP

The slope at point D is θD=2Pl2EI_.

Explanation of Solution

Given information:

The flexural rigidity of the rod is EI.

Calculation:

Calculate the slope at D (θD) as shown below.

Differentiate both side of the Equation (1) with respect to MD as shown below.

θD=UMD=12EI(0l2MABMABMDdx+0l2MBCMBCMDdx+0L22MCDMCDMDdv)=1EI(0lMABMABMDdx+0lMBCMBCMDdx+0lMCDMCDMDdv)

Substitute MD+Py for MAB, 1 for MABP, MD+Pl for MBC, 1 for MBCMD, MD+Py for MCD, and 1 for MCDMD and apply the limits in Equation (2).

θD=1EI(0l(MD+Py)(1)dy+0l(MD+Pl)(1)dx+0l(MD+Py)(1)dy)

Substitute 0 for MD.

θD=1EI(0lPydy+0lPldx+0lPydy)=PEI((y22)0l+(lx)0l+(y22)0l)=Pl22EI(1+2+1)=2Pl2EI

Therefore, the slope of point D is θD=2Pl2EI_.

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Chapter 11 Solutions

Mechanics of Materials, 7th Edition

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