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

Integrate again to obtain the beam deflection function:
Ay EIv x
3 80 kN/m x
4 80 kN/m
0m
2m
x
4
= − − − + − 7m + C 1 x + C 2 (d)
6
24
24
Evaluate constants, using boundary conditions: For this beam, the deflection is known
at x = 0 m. Substitute the boundary condition v = 0 at x = 0 m into Equation (d) to obtain
constant C 2 :
C2 = 0
(e)
Next, substitute the boundary condition v = 0 at x = 10 m into Equation (d):
Ay
0
C
6 (10m) 3
80 kN/m
4
80 kN/m
4
= − (8 m) + (3 m) + 1(10m)
24
24
3
3
∴ (166.6667 m) A + (10m) C = 13,383.3333 kN⋅m
y
1
(f)
Finally, substitute the boundary condition dv/dx = 0 at x = 10 m into Equation (c) to obtain
Ay
0
C
2 (10m) 2
80 kN/m
3
80 kN/m
3
= − (8 m) + (3 m) +
6
6
2
2
∴ (50 m ) A + C = 6,466.6667 kN⋅m
y
1
1
(g)
EI dv
dx
Equations (f) and (g) can be solved simultaneously to compute C 1 and A y :
Now that A y is known, the reactions D y and M D can be determined from Equations (a) and (b):
D
y
= 400 kN − A = 400 kN − 153.85 kN = 246.15 kN
Ans.
y
M = A (10m) − 2,200 kN⋅ m = (153.85 kN)(10 m) − 2,200 kN⋅m
D
y
C
1
3
=−1, 225.8333 kN⋅ m; A = 153.85 kN
Ans.
=−661.50 kN⋅m
Equation (c) for the beam slope and Equation (d) for the elastic curve can now be completed:
153.85 kN
80 kN/m
80 kN/m
= x − 0m
2 − x − 2m
3 + x − 7m
3 −1, 225.8333 kN⋅ m
3
2
6
6
(h)
153.85 kN
3 80 kN/m
4 80 kN/m
4 3
EIv = x − 0m − x − 2m + x − 7m − (1,225.8333 kN⋅ m) x
6
24
24
(i)
(b) Beam Deflection at C
From Equation (i), the beam deflection at C (x = 7 m) is computed as follows:
y
Ans.
EIv
C
153.85 kN
3
80 kN/m
4 3
= (7 m) − (5 m) − (1,225.8333 kN⋅m)(7 m)
6
24
3
=−1,869.075 kN⋅m
∴ v
C
1,869.075 kN⋅m
= −
2
120,000 kN⋅m
3
=− 0.015576 m = 15.58 mm ↓ Ans.
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