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1 - Shear and Moment in Beams Cest 412: Theory of Structures 1 Engr. S. Olarte

The document discusses shear and moment in beams. It explains that beams experience internal forces of shear force and bending moment when subjected to transverse loading. Shear force and bending moment vary along the length of the beam and give rise to normal and shear stresses. Knowing the distribution of shear force and bending moment is important for computing stresses and deformations in beams. The document also provides an example problem, showing how to calculate reactions, write shear and moment equations for different segments of a beam under a given loading, and draw the corresponding shear and moment diagrams.

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Sam Olarte
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247 views7 pages

1 - Shear and Moment in Beams Cest 412: Theory of Structures 1 Engr. S. Olarte

The document discusses shear and moment in beams. It explains that beams experience internal forces of shear force and bending moment when subjected to transverse loading. Shear force and bending moment vary along the length of the beam and give rise to normal and shear stresses. Knowing the distribution of shear force and bending moment is important for computing stresses and deformations in beams. The document also provides an example problem, showing how to calculate reactions, write shear and moment equations for different segments of a beam under a given loading, and draw the corresponding shear and moment diagrams.

Uploaded by

Sam Olarte
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Shear and Moment in Beams

Beam refers to a slender bar parallel to the ground that carries transverse loading, which is the applied force
perpendicular to it or its self-weight. In a beam, the internal force system consist of a shear force and a
bending moment acting on the cross section of the bar. The shear force and the bending moment usually
vary continuously along the length of the beam.

The internal forces give rise to two kinds of stresses on a transverse section of a beam: (1) normal stress
that is caused by bending moment and (2) shear stress due to the shear force. Knowing the distribution of
the shear force and the bending moment in a beam is essential for the computation of stresses and
deformations for our future discussions.

Consider a simple beam shown of length L that carries a uniform load of w (N/m) throughout its length and
is held in equilibrium by reactions R1 and R2 (which we already discussed on the previous topics).

Assume that the beam is cut at point C a distance of x from he left support and the portion of the beam to
the right of C be removed. The portion removed must then be replaced by vertical shearing force V together
with a couple M to hold the left portion of the bar in equilibrium under the action of R1 and wx.

The couple M is called the resisting moment or moment and the force V is called the resisting shear or
shear. The sign of V and M are taken to be positive if they have the senses indicated below.

1 | SHEAR AND MOMENT IN BEAMS CEST 412: THEORY OF STRUCTURES 1


ENGR. S. OLARTE
Situation 1:

Write shear and moment equations for the beam shown. In each problem, let x be the distance measured
from left end of the beam. Also, draw shear and moment diagrams, specifying values at all change of loading
positions and at points of zero shear. Neglect the mass of the beam.

a. Solve for the reactions at each supports

ΣMB=0
5RD+1(30)=3(50))
RD=24kN

ΣMD=0
5RB=2(50)+6(30)
RB=56kN

b. Divide the beam into segments at every change of loading and write the shear and moment
equation for each segment.

Segment AB:
VAB=−30kN
MAB=−30xkN⋅m

2 | SHEAR AND MOMENT IN BEAMS CEST 412: THEORY OF STRUCTURES 1


ENGR. S. OLARTE
Segment BC:
VBC=−30+56
VBC=26kN
MBC=−30x+56(x−1)
MBC=26x−56kN⋅m

Segment CD:
VCD=−30+56−50
VCD=−24kN
MCD=−30x+56(x−1)−50(x−4)
MCD=−30x+56x−56−50x+200
MCD=−24x+144kN⋅m

To draw the Shear Diagram:

1. In segment AB, the shear is uniformly distributed over the segment at a magnitude of -30 kN.

2. In segment BC, the shear is uniformly distributed at a magnitude of 26 kN.

3. In segment CD, the shear is uniformly distributed at a magnitude of -24 kN.

To draw the Moment Diagram:

3 | SHEAR AND MOMENT IN BEAMS CEST 412: THEORY OF STRUCTURES 1


ENGR. S. OLARTE
Have the value of x as the maximum possible value of it.
1. The equation MAB = -30x is linear, at x = 0, MAB = 0 and at x = 1 m, MAB = -30 kN·m.
2. MBC = 26x - 56 is also linear. At x = 1 m, MBC = -30 kN·m; at x = 4 m, MBC = 48 kN·m. When
MBC = 0, x = 2.154 m, thus the moment is zero at 1.154 m from B.
3. MCD = -24x + 144 is again linear. At x = 4 m, MCD = 48 kN·m; at x = 6 m, MCD = 0.

To check your diagram observe the following parameters.:

1. Summation of force is zero.


2. Shear diagram ends with zero.
3. Moment diagram starts at zero and ends with zero.
4. The maximum value of moment is located at zero shear.

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ENGR. S. OLARTE
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ENGR. S. OLARTE
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ENGR. S. OLARTE
7 | SHEAR AND MOMENT IN BEAMS CEST 412: THEORY OF STRUCTURES 1
ENGR. S. OLARTE

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