Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

Shear-force equation: Using the integration rules given in Equation (7.11), integrate
Equation (f) to derive the shear-force equation for the beam:
∫
V( x) = wx ( ) dx
0 − 1 0
= 73.75 kN 〈 x − 0m〉 −120 kN⋅ m〈 x − 2 m〉 − 45kN〈 x − 4 m〉
(g)
− 30 kN/m 〈 x − 6m〉 + 30 kN/m 〈 x − 9m〉 + 61.25 kN 〈 x − 12 m〉
1 1 0
Bending-moment equation: Similarly, integrate Equation (g) to derive the bendingmoment
equation for the beam:
∫
Mx ( ) = V( x)
dx
= 73.75 kN〈 x − 0m〉 −120 kN⋅ m〈 x − 2 m〉 − 45kN〈 x − 4 m〉
1 0 1
30 kN/m
30 kN/m
− 〈 x − 6m〉 + 〈 x − 9m〉 + 61.25 kN 〈 x − 12 m〉
2
2
2 2 1
(h)
A
y
120 kN.m
B
45 kN
C
2 m 2 m 2 m 3 m 3 m
73.75 kN 61.25 kN
D
30 kN/m
E
F
x
Plot the Functions
Plot the V(x) and M(x) functions given in Equations (g) and
(h) for 0 m ≤ x ≤ 12 m to create the shear-force and bendingmoment
diagram shown.
73.75 kN
73.75 kN
28.75 kN
V
6.96 m
246.28 kN.m
– 61.25 kN
– 61.25 kN
147.50 kN.m
232.50 kN.m
175.00 kN.m
183.75 kN.m
M
27.50 kN.m
ExAmpLE 7.11
y
50 kN/m
Use discontinuity functions to express the linearly distributed load acting on the
beam between A and B.
A B C
2.5 m 4.0 m
x
Plan the Solution
The expressions found in Table 7.2 are explained by means of the example of the
loading shown on the beam to the left.
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