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

SolutioN
(a) Support Reactions
An FBD of the beam is shown. Since no forces act in the
x direction, the ΣF x equation will be omitted here. From the
FBD, the beam reaction forces can be expressed by the
following relationships:
v
80 kN/m
A B C D
2 m
5 m
3 m
MD
x
A y
D y
Σ F = A + D − (80 kN/m)(5 m) = 0 ∴ A + D = 400 kN (a)
y y y y y
Σ M = − A (10m) + (80 kN/m)(5 m)(5.5 m) + M = 0
D y D
∴ M − A (10m) =−2,200 kN⋅m
D
y
(b)
Discontinuity Expressions
Distributed load between B and C: Use case 5 of Table 7.2 to write the following
expression for the distributed load:
0 0
wx ( ) =−80 kN/m x − 2m + 80 kN/m x − 7m
Reaction forces A y , D y , and M A : Since the beam is statically indeterminate, the reaction
forces at A and D can be expressed only as unknown quantities at this time:
−1 −1 −2
y y D
wx ( ) = A x − 0m + D x −10 m − M x −10m
Note that the terms for D y and M D will always have the value zero in this example, since
the beam is only 10 m long; therefore, these terms may be omitted here.
Integrate the beam load expression: The complete load expression w(x) for the beam
is thus
wx ( ) = A x − 0m − 80 kN/m x − 2m + 80 kN/m x − 7m
y
−1 0 0
The function w(x) will be integrated to obtain the shear-force function V(x):
∫
V( x) = wx ( ) dx = A x − 0m − 80 kN/m x − 2m + 80 kN/m x − 7m
EI d 2
v
2
dx
y
0 1 1
Note that a constant of integration is not needed here, since the unknown reaction at A
has been included in the function. The shear-force function is integrated to obtain the
bending-moment function M(x):
80 kN/m
80 kN/m
Mx ( ) = ∫V( x) dx = Ay
x − 0m − x − 2m + x − 7m
2
2
As before, a constant of integration is not needed for this result. However, the next two
integrations (which will produce functions for the beam slope and deflection) will require
constants of integration that must be evaluated from the beam boundary conditions.
From Equation (10.1), we can write
Now integrate the moment function to obtain an expression for the beam slope:
EI dv
dx
1 2 2
80 kN/m
80 kN/m
= Mx ( ) = Ay
x − 0m − x − 2m + x − 7m
2
2
1 2 2
Ay x
2 80 kN/m
0m
x
3 80 kN/m
2m
x
3
= − − − + − 7m + C 1 (c)
2
6
6
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