EBK MECHANICS OF MATERIALS
EBK MECHANICS OF MATERIALS
7th Edition
ISBN: 9780100257061
Author: BEER
Publisher: YUZU
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Chapter 9.6, Problem 152P
To determine

The reaction (RA), (RB) and (Rc) at each support.

Expert Solution & Answer
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Answer to Problem 152P

The reaction (RA) at the support A is 3P32(downward)_.

The reaction (RB) at the support B is 1332P(upward)_.

The reaction (RC) at the support C is 1116P(upward)_.

Explanation of Solution

Calculation:

Let, choose reaction RC as redundant.

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

EBK MECHANICS OF MATERIALS, Chapter 9.6, Problem 152P , additional homework tip  1

Calculate the vertical reaction (RB1) at point B by taking moment about A.

MA=0RB1×(2L)+RC(L)=02RB1L+RCL=02RB1L+RCL=RC22RB1L+RCL=RC2

Calculate the vertical reaction (RA1) at point A by resolving the vertical component of forces.

Fy=0RA1+RCRC2=0RA1=RC2

Show the free body diagram of the beam in below Figure 2.

EBK MECHANICS OF MATERIALS, Chapter 9.6, Problem 152P , additional homework tip  2

Calculate the vertical reaction (RB2) at point B by taking moment about A.

MA=0RB2×(2L)P(32L)=0RB2×(2L)=P(32L)RB2=(3PL2)2LRB2=3P4

Calculate the vertical reaction (RA2) at point A by resolving the vertical component of forces.

Fy=0RA2P+3P4=0RA2P4=0RA2=P4

Calculate the moment (M1) due to reaction RA2:

M1=3P4×L2=3PL8

Calculate the (M1EI) value:

Substitute 3PL8 for M1.

M1EI=(3PL8)EI=3PL8EI

Show the M1EI diagram in below Figure 3.

EBK MECHANICS OF MATERIALS, Chapter 9.6, Problem 152P , additional homework tip  3

Calculate the area (A1) using the relation:

A1=12(L+L2)(M1EI)

Substitute 3PL8EI for (M1EI).

A1=12(L+L2)(3PL8EI)=12(2L3)(3PL8EI)=6PL248EI=PL28EI

Calculate the area (A1+A2) using the relation:

(A1+A2)=12(L+L2)(M2EI)

Substitute 3PL8EI for (M2EI).

(A1+A2)=12(L+L2)(3PL8EI)=9PL232EI

Calculate the moment (M2) due to reaction RB2:

M2=RB2×(L+L2)

Substitute P4 for (M1EI).

M2=P4×(L+L2)=3PL8

Calculate the M2EI value:

Substitute 3PL8 for M2.

M2EI=3PL8EI=3PL8EI

Show the M2EI diagram in below Figure 4.

EBK MECHANICS OF MATERIALS, Chapter 9.6, Problem 152P , additional homework tip  4

Calculate the area (A3) using the relation:

(A3)=12(L2)(M2EI)

Substitute 3PL8EI for (M2EI).

(A3)=12(L2)(3PL8EI)=3PL232EI

Calculate the moment (M3) to reaction RA1:

M3=RC2×L=RCL2

Calculate the M3EI value:

Substitute (RCL2) for M3.

M3EI=(RCL2)EI=RCL2EI

Show the M3EI diagram in below Figure 5.

EBK MECHANICS OF MATERIALS, Chapter 9.6, Problem 152P , additional homework tip  5

Calculate the area (A4) using the relation:

A4=12L(M3EI)

Substitute RCL2EI for (M3EI).

A4=12L(RCL2EI)=RCL24EI

Calculate the area (A4+A5) using the relation:

(A4+A5)=12(2L)(M3EI)

Substitute RCL2EI for (M3EI).

(A4+A5)=12(2L)(RCL2EI)=RCL22EI

Calculate the tangential deviation of A with respect to B using the relation:

tB/A=(A1+A2)((13(L+L2))+L2)+A3(23L2)+(A4+A5)(122L)

Substitute 9PL232EI for (A1+A2), 3PL232EI for A3 and (RCL22EI) for (A4+A5).

tB/A=(9PL232EI)((13(L+L2))+L2)+(3PL232EI)(23L2)+(RCL22EI)(122L)=9PL332EI+PL332EIRCL22EI

Calculate the tangential deviation of C with respect to A using the relation:

tC/A=A1(13L)+A4(13L)

Substitute PL28EI for A1 and RCL24EI for A2.

tC/A=PL28EI(13L)RCL24EI(13L)=PL324EIRCL312EI

Express the deflection yC equation as below;

yC=yA+LθA+tB/A0=0+LθA+tC/AθA=tC/AL

Express the deflection yC equation as below;

yB=2LθA+tB/A

Substitute tC/AL for θA.

yB=2LtC/AL+tB/A0=2tC/A+tB/A2tC/A+tB/A=0 (1).

Calculate the reaction (RB) at the point B using the Equation (1).

2LtC/AL+tB/A=0

Substitute 9PL332EI+PL332EIRCL22EI for tB/A and PL324EIRCL312EI for tC/A.

2(PL324EIRCL312EI)+(9PL332EI+PL332EIRCL22EI)=0PL312EI+RCL36EI+9PL332EI+PL332EIRCL22EI=011PL348EIRCL33EI=0

RCL33EI=11PL348EIRC=11PL348EI×3EIL3RC=1116P

Calculate the reaction (RA) at the point A by taking the moment about B.

MB=02LRALRC+L2P=0

Substitute 1116P for RC.

2LRAL(1116P)+L2P=02LRA3PL16=02LRA=3PL16

RA=(3PL16)2LRA=332PRA=332P(downward)

Calculate the reaction (RB) at the point B by taking the moment about A.

MA=02LRB+LRC3L2P=0

Substitute 1116P for RC.

2LRB+L(1116P)3L2P=02LRB=13PL16RB=(13PL16)2LRB=1332P

Thus, the reaction (RA) at the support A is 3P32(downward)_.

Thus, the reaction (RB) at the support B is 1332P(upward)_.

Thus, the reaction (RC) at the support C is 1116P(upward)_.

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EBK MECHANICS OF MATERIALS

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