Mechanics of Materials, CE3110
Mechanics of Materials, CE3110
7th Edition
ISBN: 9781260078084
Author: N/A
Publisher: McGraw Hill Create
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Textbook Question
Chapter 9, Problem 157RP

For the loading shown, determine (a) the equation of the elastic curve for the cantilever beam AB, (b) the deflection at the free end, (c) the slope at the free end.

Chapter 9, Problem 157RP, For the loading shown, determine (a) the equation of the elastic curve for the cantilever beam AB,

Fig. P9.157

(a)

Expert Solution
Check Mark
To determine

The equation of elastic curve (y) for the cantilever beam AB.

Answer to Problem 157RP

The equation of the elastic curve for the cantilever beam AB is w0EIL(L3x26L2x312+x5120)_.

Explanation of Solution

Calculation:

Show the free body diagram of the beam as in Figure 1.

Mechanics of Materials, CE3110, Chapter 9, Problem 157RP , additional homework tip  1

Calculate the reaction (RA) at point A by resolving the vertical component of force.

Fy=0RAw0L2=0RA=w0L2

Calculate the moment MA at point A as below:

MA=0MAw0L2×2L3=0MA=w0L23MA=w0L23

Show the free body diagram of beam section AE as in Figure 2.

Mechanics of Materials, CE3110, Chapter 9, Problem 157RP , additional homework tip  2

Calculate the moment (M) by taking moment about the point E.

ME=0w0L23w0Lx2+w0x22L×(x3)+M=0M=w0L23+w0Lx2w0x22L×(x3)M=w0L23+w0Lx2w0x36LM=w0L23+w0Lx2w0x36L (1).

Calculate the equation of the elastic curve (y) by integrating the Equation (1).

EId2ydx2=w0L23+w0Lx2w0x36LEIdydx=w0L23x+w0L2x22w06Lx44+C1EIdydx=w0L23x+w0Lx24w0x424L+C1 (2).

Substitute 0 for x and 0 for dydx in Equation (2).

EI(0)=w0L23(0)+w0L(0)24w0(0)424L+C10=0+00+C1C1=0

Integrate equation (2) and Substitute 0 for C1.

EIdydx=w0L23x+w0Lx24w0x424L+(0)EIdydx=w0L23x22+w0L4x33w024Lx55+C2EIy=w0L2x26+w0Lx312w0x5120L+C2 (3)

Substitute 0 for x and 0 for y in Equation (3).

EI(0)=w0L2(0)26+w0L(0)312w0(0)5120L+C20=0+00+C2C2=0

Integrate Equation (3) and Substitute 0 for C2.

EIy=w0L2x26+w0Lx312w0x5120L+0

Multiply the Equation by (LL).

EIy=w0L2x26LL+w0Lx312LLw0x5120LLLEIy=w0L3x26L+w0L2x312Lw0x5120Ly=w0EIL(L3x26L2x312+x5120)

Thus, the equation of the elastic curve for the cantilever beam AB is w0EIL(L3x26L2x312+x5120)_.

(b)

Expert Solution
Check Mark
To determine

The deflection (yB) the free end at point B.

Answer to Problem 157RP

The deflection (yB) the free end at point B is 1112w0L4EI(downward)_.

Explanation of Solution

Calculation:

Calculate the deflection (yB) using the Equation (3).

Substitute L for x.

EIy=w0L2(L)26+w0L(L)312w0(L)5120LEIy=w0L46+w0L412w0(L)5120LyB=w0L4EI(16112+1120)yB=11120w0L4EI

Thus, the deflection (yB) the free end at point B is 1112w0L4EI(downward)_.

(c)

Expert Solution
Check Mark
To determine

The slope (θB) the free end at point B.

Answer to Problem 157RP

The slope (θB) the free end at point B is 18w0L3EI_.

Explanation of Solution

Calculation:

Calculate the slope (θB) using the Equation (2).

Substitute L for x and 0 for C1

EIdydx=w0L23(L)+w0L(L)24w0(L)424L+0EIdydx=w0L33+w0L34w0L324dydx=w0L3EI(1314+124)dydx=18w0L3EI

Thus, the slope (θB) the free end at point B is 18w0L3EI_.

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Chapter 9 Solutions

Mechanics of Materials, CE3110

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