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

The inclusion of the reaction forces in the expression for w(x) has automatically accounted
for the constants of integration 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
EI d 2
v
0 1
= Mx ( ) = −224 kip⋅ft x − 0 ft + 24 kips x − 0ft
2
dx
6kips/ft
6kips/ft
6kips/ft
− x − 4ft + x − 12 ft + x −12 ft
6(8 ft)
6(8 ft)
2
3 3 2
Integrate the moment function to obtain an expression for the beam slope:
EI dv
dx
1
24 kips
2
=−224 kip⋅ft〈 x − 0 ft〉 + 〈 x − 0ft〉
2
6kips/ft
4
6kips/ft
4
6kips/ft
3
− 〈 x − 4ft〉 + 〈 x − 12 ft〉 + 〈 x − 12 ft〉
+ C
24(8 ft)
24(8 ft)
6
1
(a)
Integrate again to obtain the beam deflection function:
224 kip⋅ft
2 24 kips
3
EIv =− x − 0ft + x − 0ft
2
6
6kips/ft
5 6kips/ft
5 6kips/ft
4
− x − 4ft + x − 12 ft + x − 12 ft + C x + C
120(8 ft)
120(8 ft)
24
1 2
(b)
EIv
D
224 kip⋅ft
=− (16ft)
2
=−13,772.8 kip⋅ft
Evaluate the constants, using boundary conditions: For this beam, the slope and the deflection
are known at x = 0 ft. Substitute the boundary condition dv/dx = 0 at x = 0 ft into
Equation (a) to obtain
C1
= 0
Next, substitute the boundary condition v = 0 at x = 0 ft into Equation (b) to obtain the
constant C 2 :
C2
= 0
The beam slope and elastic curve equations are now complete:
EI dv
1 24 kips
2
=−224 kip⋅ft x − 0 ft + 〈 x − 0ft〉
dx
2
6kips/ft
6kips/ft
6kips/ft
− x − 4ft + 〈 x − 12 ft〉 + 〈 x −12 ft〉
24(8 ft)
24(8 ft)
6
224 kip⋅ft
2 24 kips
3
EIv =− x − 0ft + 〈 x − 0ft〉
2
6
6kips/ft
6kips/ft
6kips/ft
− x − 4ft + x − 12 ft + x −12 ft
120(8 ft)
120(8 ft)
24
4 4 3
5 5 4
Beam deflection at D: The beam deflection at D (where x = 16 ft) is computed as follows:
3
∴ v
24 kips 6kips/ft
+ (16ft) − + +
6
120(8 ft) (12ft) 6kips/ft
120(8 ft) (4 ft) 6kips/ft
(4 ft)
24
2 3 5 5 4
D
13,772.8 kip⋅ft
= −
192,000 kip⋅ft
3
2
=− 0.071733 ft = 0.861 in. ↓
Ans.
420