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

9 kips/ft
5 kips/ft
11 kips/ft
Bending-moment equation: Similarly, integrate Equation (g)
to derive the bending-moment equation for the beam:
V
A
B
6 ft 6 ft 6 ft 6 ft
29 kips
– 27 kips
66 kip·ft
C
56 kips 19 kips
29 kips
4.11 ft
D
–19 kips
131.39 kip·ft
114 kip·ft
–19 kips
E
x
∫
M( x) = V( x)
dx
9kips/ft
9kips/ft
=− 〈 x − 0ft〉 + 〈 x − 0ft〉
2
6(6ft)
2 3
9kips/ft
− 〈 x − 6ft〉 + 56.0 kips 〈 x − 6ft〉
66ft ( )
3 1
5kips/ft
5kips/ft
− 〈 x − 12 ft〉 + 〈 x − 18 ft〉
2
2
6kips/ft
6kips/ft
− 〈 x − 12 ft〉 + 〈 x − 18 ft〉
66ft ( )
6(6ft)
2 2
3 3
6kips/ft
+ 〈 x − 18 ft〉 + 19 kips〈 x − 24 ft〉
2
2 1
(h)
M
–108 kip·ft
Plot the Functions
Plot the V(x) and M(x) functions given in Equations (g) and (h)
for 0 ft ≤ x ≤ 24 ft to create the shear-force and bending-moment
diagram shown.
pRoBLEmS
p7.32–p7.42 For the beams and loadings shown in Figures
P7.32–P7.42,
(a) use discontinuity functions to write the expression for w(x);
include the beam reactions in this expression.
(b) integrate w(x) twice to determine V(x) and M(x).
(c) use V(x) and M(x) to plot the shear-force and bending-moment
diagrams.
A
10 kN 35 kN
B
C
2.5 m
3 m
2 m
D
x
FIGURE p7.33
y
180 lb
450 lb
y
30 kN 20 kN
15 kN
x
x
A
B
2 ft 4 ft
C
3 ft
D
A
B C D
3 m 4 m
3 m
5 m
E
FIGURE p7.32
FIGURE p7.34
234