Mechanics of Materials (MindTap Course List)
9th Edition
ISBN: 9781337093347
Author: Barry J. Goodno, James M. Gere
Publisher: Cengage Learning
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Textbook Question
Chapter 4, Problem 4.3.20P
Find expressions for shear force V and moment M at x = L/2 of beam BC. Express V and M in term s of peak load intensity q0and be a m length variable L.
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For the beam shown, derive the expressions for shear and moment. Show complete solutions. Simplify
all equations. Then draw the shear force and bending moment diagrams below the load diagram.
Draw to scale.
1. Shear and Moment Diagram:
18 kN/m
25 kN-m
A
E
|C
ImIm
- 2 m
3 m·
Shear Equation
Moment Equation
Segment AB
Segment BC
Segment CD
Segment DE
Point of Zero Shear:
Moment at point of zero shear:
The simply supported beam is subjected to the force F = 700 N and the uniform distributed load with
intensity w = 150 N/m. Draw the shear force and bending moment diagrams (in your homework
documentation) and determine the equations for V(r) and M(x). Take a = 0 at point A.
19
F
a
Values for dimensions on the figure are given in the following table. Note the figure may not be to scale.
Variable Value
a
5.2 m
2.6 m
3.12 m
Support Reactions
The reaction at A is
N.
The reaction at D is
N.
Shear Force and Bending Moment Equations
In section AB:
V(x)=
N and M(x)=
N-m.
In section BC:
v(x)-
N and M(x)=
N-m.
In section CD:
V(x)-
N and M(x)=
N-m.
A
3
For the beam shown, find the reactions at the supports and plot the shear-force and bending-moment diagrams. V
= 9 kN, V2 = 9 kN, V3 = 200 mm, and V4 = 1100 mm.
ATAT-V3
Provide values at all key points shown in the given shear-force and bending-moment diagrams.
X
(mm)
B
A =
B =
C =
D =
E=
F=
P =
Q =
E
* KN
* KN
* KN
× KN
KN
x KN
✩ kN.mm
*kN.mm
D
0.00
Reaction force R₁ (left) =
In the shear-force and bending-moment diagrams given,
+V
0.00
X
(mm)
6.3 kN and reaction force R2 (right) =
P
11.7 kN.
Q
0.00
Chapter 4 Solutions
Mechanics of Materials (MindTap Course List)
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