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

(a) Beam Deflection at A
At the tip of the overhang, where x = 0 m, the beam deflection is
25 kN
35 kN
EIvA
=− x − 0m +
6
2
x − 2m + (133.3333 kN⋅m) x − 233.3333 kN⋅m
3
=−233.3333 kN⋅m
∴ v
A
(b) Beam Deflection at C
At C, where x = 4.5 m, the beam deflection is
3 3 2 3
233.3333 kN⋅m
= −
17 × 10 kN⋅m
3
3 2
=− 0.013725 m = 13.73 mm ↓
Ans.
EIv
C
25 kN 35 kN
=− (4.5 m) + (2.5 m) + (133.3333 kN⋅m)(4.5 m) − 233.3333 kN⋅m
6
6
3
= 78.1249 kN⋅m
3 3 2 3
78.1249 kN⋅m
∴ vC
=
17 × 10 kN⋅m
3
3 2
= 0.004596 m = 4.60 mm ↑
Ans.
ExAmpLE 10.7
v
60 kN/m
40 kN/m
x
For the beam shown, use discontinuity functions to compute
(a) the slope of the beam at A.
(b) the deflection of the beam at B.
A
v
A
A y
B
4 m 5 m 3 m 3 m
60 kN/m
B
C
C
D
40 kN/m
4 m 5 m 3 m 3 m
D
D y
E
E
x
Assume a constant value of EI = 125 × 10 3 kN · m 2 for the beam.
Plan the Solution
Determine the reactions at simple supports A and D. Using
Table 7.2, write w(x) expressions for the two uniformly distributed
loads as well as the two support reactions. Integrate
w(x) four times to determine equations for the beam slope and
deflection. Use the boundary conditions known at the simple
supports to evaluate the constants of integration.
SolutioN
Support Reactions
An FBD of the beam is shown. From this FBD, the beam
reaction forces can be computed as follows:
Σ M = −(60 kN/m)(4 m)(2 m) − (40 kN/m)(6 m)(12m) + D (12 m) = 0
A
∴ D = 280 kN
y
Σ F = A + D − (60 kN/m)(4 m) − (40 kN/m)(6 m) = 0
y y y
∴ A = 200 kN
y
Discontinuity Expressions
Distributed load between A and B: Use case 5 of Table 7.2 to write the following
expression for the 60 kN/m distributed load:
0 0
wx ( ) =−60 kN/m x − 0m + 60 kN/m x − 4m
y
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