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

Integrate again to obtain the beam deflection function:
200 kN
60 kN/m
60 kN/m
EIv = x − 0m − x − 0m + x − 4m
6
24
24
40 kN/m
4 280 kN
3
− x − 9m + x − 12 m + Cx 1 + C2
24
6
3 4 4
(b)
Evaluate the constants, using boundary conditions: Boundary conditions are specific values
of the deflection v or slope dv/dx that are known at particular locations along the beam span.
For this beam, the deflection v is known at the pin support (x = 0 m) and at the roller support
(x = 12 m). Substitute the boundary condition v = 0 at x = 0 m into Equation (b) to obtain
C2
= 0
Next, substitute the boundary condition v = 0 at x = 12 m into Equation (b) to obtain the
constant C 1 :
200 kN
6
(12m)
60 kN/m 60 kN/m 40 kN/m
− (12m) + (8 m) − (3 m) + C1(12m) = 0
24
24
24
2
∴ C = −1,322.0833 kN⋅m
3 4 4 4
The beam slope and elastic curve equations are now complete:
EI dv
dx
1
200 kN
2 60 kN/m
3 60 kN/m
3
= x − 0m − x − 0m + x − 4m
2
6
6
40 kN/m
280 kN
− x − 9m + x −12 m −1,322.0833 kN⋅m
6
2
3 2 2
200 kN
3 60 kN/m
4 60 kN/m
4
EIv = x − 0m − x − 0m + x − 4m
6
24
24
40 kN/m
4 280 kN
3 2
− x − 9m + x −12 m − (1,322.0833 kN⋅m)
x
24
6
(a) Beam Slope at A
The beam slope at A (x = 0 m) is
⎛ dv ⎞
EI
⎝
⎜
⎠
⎟ =−1,322.0833 kN⋅m
dx
∴ ⎛ ⎝ ⎜
dv ⎞ 1,322.0833 kN⋅m
⎠
⎟ =−
3 2
dx 125 × 10 kN⋅m
A
A
2
2
=−0.01058 rad
Ans.
(b) Beam Deflection at B
The beam deflection at B (x = 4 m) is
EIv
B
200 kN
3
60 kN/m
4 2 3
= (4 m) − (4 m) − (1,322.0833 kN⋅ m)(4 m) =−3,795 kN⋅m
6
24
3
3,795 kN⋅m
∴ vB
= −
=− 0.030360 m = 30.4 mm ↓ Ans.
3 2
125 × 10 kN⋅m
418