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Shear and Moment Computation

The document discusses determining internal forces in beams. It provides an example of drawing the shear and moment diagrams for a beam subjected to a uniformly distributed load. The maximum shear is found to be -180 kN and the maximum bending moment is determined to be 288 kN-m by analyzing the free body diagram of the beam element. Equations are written to solve for the reactions and internal forces as a function of position along the beam.

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78 views4 pages

Shear and Moment Computation

The document discusses determining internal forces in beams. It provides an example of drawing the shear and moment diagrams for a beam subjected to a uniformly distributed load. The maximum shear is found to be -180 kN and the maximum bending moment is determined to be 288 kN-m by analyzing the free body diagram of the beam element. Equations are written to solve for the reactions and internal forces as a function of position along the beam.

Uploaded by

jen
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Solving For Internal Forces:

1. Draw the Shear and


Moment Diagram.
2. Determine Maximum
Shear and Bending Moments
on Beams
Sample Beam Element:
Example:
1. Draw the Shear and Moment Diagram
2. Determine the maximum shear and 36 kN/m

bending moment of the beam shown.


a b c
9m 3m

=0
Ra Rb
36(12)(6) – Rb(9) = 0
Rb = 288 kN

=0

36(12) – (Ra + Rb) = 0


Ra = 144 kN
Example:
1. Draw the Shear and Moment Diagram
2. Determine the maximum shear and 36 kN/m

bending moment of the beam shown.


a b c
9m 3m
Shear
1) 144 kN 144 – 36(X) = 0
2) 144 – 36(9) = -180 kN (Vmax) Ra Rb
X= 4
3) -180 + 288 = 108 kN
4) 108 – 36(3) = 0 kN 144 108
0.5(144)(X)
0.5(108)(3) 0
Moment 288 0.5(-180)(9-X)

2) ½(144)(X) = 288 kN-m


-180
3) 288 + ½(-180)(9-X) = -162 kN-m
0 0
4) -162 + ½(108)(3) = 0 kN-m

-162

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