Fundamentals Of Structural Analysis:
Fundamentals Of Structural Analysis:
5th Edition
ISBN: 9781260083330
Author: Leet, Kenneth
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 13, Problem 2P
To determine

Find the moment at the supports B and C of the beam using approximate analysis.

Sketch the shear and moment diagrams of the beam.

Expert Solution & Answer
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Answer to Problem 2P

Case 1: Length L1=3m.

The moment at the supports B and C of the beam using approximate analysis are 12.86kNm_ and 16.88kNm_.

The moment at the supports B and C of the beam using moment distribution method are 13.5kNm_ and 15.75kNm_.

Case 2: Length L1=12m.

The moment at the supports B and C of the beam using approximate analysis are 24.36kNm_ and 6.3kNm_.

The moment at the supports B and C of the beam using moment distribution method are 36.84kNm_ and 4.1kNm_.

Explanation of Solution

Given information:

The value of EI is constant.

Case 1: Length L1=3m.

Case 2: Length L1=12m.

Apply the sign conventions for calculating reactions, forces, and moments using the three equations of equilibrium as shown below.

  • For summation of forces along x-direction is equal to zero (Fx=0), consider the forces acting towards right side as positive (+) and the forces acting towards left side as negative ().
  • For summation of forces along y-direction is equal to zero (Fy=0), consider the upward force as positive (+) and the downward force as negative ().
  • For summation of moment about a point is equal to zero (Matapoint=0), consider the clockwise moment as negative and the counter clockwise moment as positive.

Calculation:

Case 1: Length L1=3m.

For span AB,

Let consider the location of a point of inflection at 0.3L to the left of support B which means 0.9 m to the left of support B or 2.1 m from the support A to the right.

For span BC,

Let consider the location of a point of inflection at 0.23L to the left of support C which means 1.38 m to the left of support B or 4.62 m from the support B to the right.

Show the location of a point of inflection of span AB and BC as in Figure 1.

Fundamentals Of Structural Analysis:, Chapter 13, Problem 2P , additional homework tip  1

Refer to Figure 1.

Consider P.I. of span AB,

Find the reaction RA at support A:

Summation of moments about P.I. is equal to 0.

MP.I=0RA(2.1)20(0.6)=0RA=5.71kN

Consider P.I. of span BC,

Find the reaction RB at support B:

Summation of moments about P.I. is equal to 0.

MP.I=014.29(5.52)+RB(4.62)23.1(2.31)=0RB=28.62kN

Consider entire structure,

Find the reaction RC at support C:

Summation of moments about P.I. is equal to 0.

MP.I=05.7120+28.6230+RC=0RC=15.67kN

Find the shear values of the beam:

FA=5.71kN

Fat20kN,JL=5.71kN

Fat20kN,JR=5.7120=14.29kN

FB,JL=5.7120=14.29kN

FB,JR=5.7120+28.62=14.33kN

FC=15.67kN

Find the point of zero shear force for span BC,

Fxx=015.675x=0x=3.134mfrompointC

Find the moment values of the beam:

MA=0

Mat20kNload=5.71(1.5)=8.57kNm

MB=5.71(3)20(1.5)=12.87kNm

MC=16.88kNm

Mx=3.134mfrompointC=15.67(3.134)5(3.134)(3.1342)16.88=7.67kNm

Sketch the shear and moment diagram of the beam using approximate analysis as shown in Figure 2.

Fundamentals Of Structural Analysis:, Chapter 13, Problem 2P , additional homework tip  2

Check the shear and moment values using moment distribution method:

Find the fixed end moments in the span AB and BC:

FEMAB=PL8=20(3)8=7.5kNm

FEMBA=FEMAB=7.5kNm

FEMBC=wL212=5(6)212=15kNm

FEMCB=FEMBC=15kNm

Find the distribution factor at joint B:

KAB=34I3=0.25I

KBC=I6=0.167I

Ks=0.25I+0.167I=0.417I

DFAB=0.25I0.417I=0.6

DFBC=0.167I0.417I=0.4

Consider span AB, find the reactions at the supports:

Summation of moments about B is equal to 0.

MB=0RA(3)20(1.5)+13.5=0RA=5.5kN

Summation of forces along y-direction is equal to 0.

+Fy=0RA+RB20=05.5+RB20=0RB=14.5kN

Consider span BC, find the reactions at the supports:

Summation of moments about B is equal to 0.

MB=0RC(6)+13.515.755(6)(62)=0RC=15.375kN

Summation of forces along y-direction is equal to 0.

+Fy=0RB+RC5(6)=0RB+15.3755(6)=0RB=14.625kN

Therefore, the total RB is 14.5+14.625=29.125kN

Find the shear values of the beam:

FA=5.5kN

Fat20kN,JL=5.5kN

Fat20kN,JR=5.520=14.5kN

FB,JL=5.520=14.5kN

FB,JR=5.520+29.125=14.625kN

FC=15.375kN

Find the point of zero shear force for span BC,

Fxx=015.3755x=0x=3.075mfrompointC

Find the moment values of the beam:

MA=0

Mat20kNload=5.5(1.5)=8.25kNm

MB=5.5(3)20(1.5)=13.5kNm

MC=15.75kNm

Mx=3.075mfrompointC=15.375(3.075)5(3.075)(3.0752)16.88=7.89kNm

Sketch the moment distribution results, shear, and moment diagrams of the beam using moment distribution method as shown in Figure 3.

Fundamentals Of Structural Analysis:, Chapter 13, Problem 2P , additional homework tip  3

Compare the results of approximate analysis and moment distribution method:

The approximate analysis results compare closely with moment distribution method.

Therefore, the moment at the supports B and C of the beam using approximate analysis are 12.86kNm_ and 16.88kNm_.

Therefore, the moment at the supports B and C of the beam using moment distribution method are 13.5kNm_ and 15.75kNm_.

Case 2: Length L1=12m.

For span AB,

Length of span AB is longer, so a larger counterclockwise moment is applied member AB to the left end of member BC. Thus the moment rotates B-end of member BC in the counterclockwise direction causes the P.I. to shift to the right.

For span BC,

Let consider the location of a point of inflection adjacent to B at 0.3L and the one on the right at 0.1L.

Show the location of a point of inflection as in Figure 4.

Fundamentals Of Structural Analysis:, Chapter 13, Problem 2P , additional homework tip  4

Sketch the reactions, forces, and moments for the segments as in Figure 5.

Fundamentals Of Structural Analysis:, Chapter 13, Problem 2P , additional homework tip  5

Refer to Figure 5.

Find the reaction RB at support B:

Summation of moments about A is equal to 0.

MA=0RB(12)20(6)9(0.9+12)9(13.8)=0RB=30.03kN

Find the shear values of the beam:

FA=7.97kN

Fat20kN,JL=7.97kN

Fat20kN,JR=7.9720=12.03kN

FB,JL=7.9720=12.03kN

FB,JR=7.9720+30.03=18kN

FC=12kN

Find the point of zero shear force for span BC,

Fxx=0125x=0x=2.4mfrompointC

Find the moment values of the beam:

MA=0

Mat20kNload=7.97(6)=47.82kNm

MB=7.97(12)20(6)=24.36kNm

MC=6.3kNm

Mx=2.4mfrompointC=12(2.4)5(2.4)(2.42)6.3=8.1kNm

Sketch the shear and moment diagram of the beam using approximate analysis as shown in Figure 6.

Fundamentals Of Structural Analysis:, Chapter 13, Problem 2P , additional homework tip  6

Check the shear and moment values using moment distribution method:

Find the fixed end moments in the span AB and BC:

FEMAB=PL8=20(12)8=30kNm

FEMBA=FEMAB=30kNm

FEMBC=wL212=5(6)212=15kNm

FEMCB=FEMBC=15kNm

Consider span AB, find the reactions at the supports:

Summation of moments about B is equal to 0.

MB=0RA(12)20(6)+36.84=0RA=6.93kN

Summation of forces along y-direction is equal to 0.

+Fy=0RA+RB20=06.93+RB20=0RB=13.07kN

Consider span BC, find the reactions at the supports:

Summation of moments about B is equal to 0.

MB=0RC(6)+36.844.15(6)(62)=0RC=9.54kN

Summation of forces along y-direction is equal to 0.

+Fy=0RB+RC5(6)=0RB+9.545(6)=0RB=20.46kN

Therefore, the total RB is 13.07+20.46=33.53kN

Find the shear values of the beam:

FA=6.93kN

Fat20kN,JL=6.93kN

Fat20kN,JR=6.9320=13.07kN

FB,JL=6.9320=13.07kN

FB,JR=6.9320+33.53=20.46kN

FC=9.54kN

Find the point of zero shear force for span BC,

Fxx=09.545x=0x=1.908mfrompointC

Find the moment values of the beam:

MA=0

Mat20kNload=6.93(6)=41.58kNm

MB=6.93(12)20(6)=36.84kNm

MC=4.1kNm

Mx=1.908mfrompointC=9.4(1.908)5(1.908)(1.9082)4.1=4.73kNm

Sketch the shear, and moment diagrams of the beam using moment distribution method as shown in Figure 7.

Fundamentals Of Structural Analysis:, Chapter 13, Problem 2P , additional homework tip  7

Compare the results of approximate analysis and moment distribution method:

Differences exist between the approximate analysis and the moment distribution method results.

The moment distribution method provides exact results.

Therefore, the moment at the supports B and C of the beam using approximate analysis are 24.36kNm_ and 6.3kNm_.

Therefore, the moment at the supports B and C of the beam using moment distribution method are 36.84kNm_ and 4.1kNm_.

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