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

ExAmpLE 7.12
9 kips/ft
A
9 kips/ft
A
5 kips/ft
12 ft
C
B y
6 ft
B
5 kips/ft
5 kips/ft
6 ft 6 ft 6 ft 6 ft
6 kips/ft
D
x
C
11 kips/ft
11 kips/ft
D
E y
E
x
Use discontinuity functions to obtain expressions for the internal
shear force V(x) and internal bending moment M(x) in the
beam shown. Then, use these expressions to plot the shear-force
and bending-moment diagrams for the beam.
B C D
E Plan the Solution
6 ft 6 ft 6 ft 6 ft
Determine the reactions at simple supports A and E. Using
Table 7.2, write w(x) expressions for the linearly decreasing
load between A and B and for the linearly increasing load between C and D, as well
as for the two support reactions. Integrate w(x) to determine an equation for the
shear force V(x), and then integrate V(x) to determine an equation for the bending
moment M(x). Plot these functions to complete the shear-force and bending-moment
diagrams.
Ex
SolutioN
Support Reactions
An FBD of the beam is shown. Before beginning, it is convenient
to subdivide the linearly increasing load between
C and D into
(a) a uniformly distributed load that has an intensity of
5 kips/ft and
(b) a linearly distributed load that has a maximum
intensity of 6 kips/ft.
Accordingly, the beam equilibrium equations are as follows:
Σ F = E = 0 (trivial)
x
x
1
F B E
2 (9 kips/ft)(6 ft) (5kips/ft)(6 ft) 1
Σ y = y + y − − − (6 kips/ft)(6 ft) = 0
2
1
Σ MB
= (9 kips/ft)(6 ft)(4 ft) − (5 kips/ft)(6 ft)(9 ft)
2
1
− (6 kips/ft)(6 ft)(10 ft) + Ey(18 ft) = 0
2
232
From these equations, the beam reactions are
B
y
= 56.0 kips and E = 19.0 kips
Discontinuity Expressions
Decreasing linearly distributed load between A and B: Use case 7 of Table 7.2 to write
the following expression for the 9 kips/ft linearly distributed loading:
0
9kips/ft
1
9kips/ft
1
wx ( ) =− 9kips/ft〈 x − 0ft〉 + 〈 x − 0ft〉 − 〈 x − 6ft〉 (a)
6ft
6ft
Reaction force B y : The upward reaction force at B is expressed with the use of case 2
of Table 7.2:
wx ( ) = 56.0 kips x − 6ft
y
−1
(b)