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

EI dv
dx
∫
Integrate the beam load expression: The complete load expression w(x) for the beam is thus
0 −1 0 −1
y
y
wx ( ) =−90 kN/m x − 0m + B x − 2m + 90 kN/m x − 12 m + D x − 12m
The function w(x) will be integrated to obtain the shear-force function V(x):
The shear-force function is integrated to obtain the bending-moment function M(x):
90 kN/m
2 1 90 kN/m
2 1
Mx ( ) = ∫V( x)
dx = − x − 0m + By
x − 2m + x − 12 m + Dy
x −12m
2
2
Since the reactions have been included in these functions, constants of integration are not
needed up to this point. 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:
90 kN/m
3 By
=− x − + x
2 90 kN/m
D
− + x
3 y
0m
2m
− + x
2
12 m − 12 m + C 1 (c)
6
2
6
2
Integrate again to obtain the beam deflection function:
1 0 1 0
y
y
V( x) = wx ( ) dx = −90 kN/m x − 0m + B x − 2m + 90 kN/m x − 12 m + D x −12m
EI d 2
v
90 kN/m
90 kN/m
= Mx ( ) = − x − 0m + B x − 2m + x − 12 m + D x −12m
2
y
y
dx
2
2
2 1 2 1
90 kN/m
4 By
EIv =− x − + x
3 90 kN/m
D
− + x
4 y
0m
− + x
3
2m
12 m − 12 m + C 1 x + C 2 (d)
24
6
24
6
Evaluate constants, using boundary conditions: For this beam, substitute the boundary
condition v = 0 at x = 2 m into Equation (d):
90 kN/m
4
0 =− (2 m) + C (2 m) + C
24
3
∴ C (2 m) + C = 60 kN⋅m
1 2
1 2
(e)
Next, substitute the boundary condition v = 0 at x = 12 m into Equation (d):
90 kN/m B
4 y 3
0 =− (12m) + (10m) + C (12 m) + C
24
6
3
∴ B (166.6667 m) + C (12m) + C = 77,760 kN⋅m
y
1 2
1 2
3
(f)
Finally, substitute the boundary condition v = 0 at x = 17 m into Equation (d):
90 kN/m B
D
4 y
0
(17m)
C C
24
6 (15m) 3
90 kN/m
4 y 3
=− + + (5 m) + (5 m) + (17m) +
24
6
3 3
3
∴ B (562.5 m) + D (20.8333 m) + C (17m) + C = 310,860 kN⋅m
y
y
1 2
1 2
(g)
458
Five equations—Equations (a), (b), (e), (f), and (g)—must be solved simultaneously to
determine the beam reaction forces at B, D, and E, as well as the two constants of integration
C 1 and C 2 :
C
1
2
=−1,880 kN⋅ m and C = 3,820 kN⋅m
B = 579 kN D = 639 kN E =−138 kN
y y y
2
3
Ans.