Example: Determine the vertical and horizontal deflections at the point B of the truss
shown.
Solution: Apply a horizontal fictitious load P1 =0 at B and replace the 84kN load by P2
so that differentiation of the total strain energy can be carried out with respect to P1 and.
P2 . P1 =0 and P2=84kN
Ax
P1
Ay
∑F
x
P2
= 0 → A x = 35 − P1
Cy
∑M
A
∑F
y
= 0 →35(4) + 7C y − 4P2 = 0 → C y =
4
P2 − 20
7
3
= 0 → A y +C y − P2 = 0 → A y = 20 + P2
7
Using method of joints, calculate member forces. Take the partial derivatives of the
member forces with respect to P1 and P2
20 +
Member
AB
BC
AD
BD
CD
3
P2
7
L(m)
4
3
5.66
4
5
−20 +
F(kN)
-15+ P1+ 0.43P2
-15+ 0.43P2
-28.28-0.61P2
P2
25-0.71P2
TOTAL
4
P2
7
δF
δ P1
δF
δ P2
1
0
0
0
0
0.43
0.43
-0.61
1
-0.71
FL
δF
δ P1
FL
P1 = 0
P2 =84
84.48
0
0
0
0
84.48
⎛
1
⎜ FL δ F
xB =
∑
δ P1
EA ⎜⎜
⎝
⎞
84.48
⎟ = 84.48 kNm =
= 0.00035m
200(106 )0.0012
AE
P1 = 0 ⎟
⎟
P2 =84 ⎠
⎛
1
⎜ FL δ F
yB =
∑
δ P2
EA ⎜⎜
⎝
⎞
797.08
⎟ = 797.08 kNm =
= 0.00332m
⎟
AE
200(106 )0.0012
P1 = 0 ⎟
P2 =84 ⎠
δF
δ P2
P1 = 0
P2 =84
36.32
37.24
274.55
336.00
122.97
797.08