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

to obtain the shear-force function V(x):
∫
0 0
V( x) = wx ( ) dx = −25 kN x − 0m + 35 kN x − 2m
Integrate again to obtain the bending-moment function M(x):
∫
1 1
Mx ( ) = V( x) dx = −25 kN x − 0m + 35 kN x − 2m
Note that, since w(x) is written in terms of both the loads and the reactions, no constants
of integration have been needed up to this point in the calculation. However, the next two
integrations (which will produce functions for the beam slope and deflection) will require
constants of integration that must be evaluated by using the beam boundary conditions.
From Equation (10.1), we can write
EI d 2
v
2
dx
1 1
= Mx ( ) = −25 kN x − 0m + 35 kN x − 2m
Integrate the moment function to obtain an expression for the beam slope:
EI dv
dx
25 kN
2 35 kN
2
=− x − 0m + x − 2m + C1 (a)
2
2
Integrate again to obtain the beam deflection function:
EIv =− 25 kN
3 35 kN
3
x − 0m + x − 2m + C1x + C2 (b)
6
6
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 roller support (x = 2 m) and at the pin
support (x = 7 m). Substitute the boundary condition v = 0 at x = 2 m into Equation (b)
to obtain
25 kN
3
35 kN
3
− (2 m) + (0 m) + C1(2 m) + C2 = 0
(c)
6
6
Next, substitute the boundary condition v = 0 at x = 7 m into Equation (b) to obtain
25 kN
3
35 kN
3
− (7 m) + (5 m) + C1(7 m) + C2 = 0
(d)
6
6
Solve Equations (c) and (d) simultaneously for the two constants of integration C 1 and C 2 :
C
1
2
= 133.3333 kN⋅ m and C =−233.3333 kN⋅m
2
3
The beam slope and elastic curve equations are now complete:
EI dv 25 kN
2 35 kN
2 2
=− x − 0m + x − 2m + 133.3333 kN⋅m
dx 2
2
25 kN
35 kN
EIv =− x − 0m + x − 2m + (133.3333 kN⋅m) x − 233.3333 kN⋅m
6
6
3 3 2 3
415