Problem: Solve for the support reactions. Determine the bar forces. A = 1000mm2 , E = 200GPa. Use the
Theorem of Least Work.
C
B
D
12m
b
a
e
d
c
50kN
Solution:
9m
9m
8m
8m
8m
8m
9m
9m
8m
8m
8m
8m
8m
8m
8m
8m
1. Check indeterminacy:
o
o
No. of unknowns: 3 (Ha , Va , Ve)
8m
8m
No. of Equilibrium equations: 3 ( ∑ 𝑀 = 0, ∑ 𝐹𝑦 = 0, ∑ 𝐹𝑥 = 0 )
: Truss is STATICALLY DETERMINATE (We can solve for all reactions)
o
Alternatively,
is b + r = 2j ?
15 + 3 ? 2(8)
18 > 16
statically INDETERMINATE TO THE 2ND ORDER.
(We cannot solve all bar forces)
>There are 2 redundant forces.
>2 compatibility equations are required
2. Primary Structure and redundant forces
o
Select bar force Cb and Cd as the redundant forces.
o
We now have a primary structure shown below equivalent to the original structure
o
X1 is the bar force at Cb; X2 is the bar force at Cd.
o
The truss in the primary structure is now statically determinate and stable with
unknown redundant forces.
Primary Structure:
C
B
D
X1
X2
12m
b
a
e
d
c
50kN
Ha
Va
9m
9m
8m
8m
9m
9m
8m
8m
8m
3. Bar Forces at Primary Structure in terms of X1 and X2:
8m
8m
8m
Express the bar forces in terms of X1 and X2
o
8m
B -18.75-0.6X1 C -18.75-0.6X2 D
8m
28.125
b
28.125-0.6X1
a
X2
9.375-0.6X2
8m
8m
8m
8m
-0.8X2
X1
X2
-0.8X1-0.8X2
50-0.8X1
X1
Ve
c
9.375
d
e
50kN
Ha = 0
Ve = 12.5
Va = 37.5
4. Solving for X1 and X2 by the Theorem of Least Work :
o
Compatibility equations are required. We obtain this from our relative axial
deformations.
o
By examining our original structure, the relative axial displacement between C and b is
0:
ΔCb or Δ1 = 0
∆1 = ∑
o
𝑃(𝜕𝑃⁄𝜕𝑋1 )𝐿
=0
𝐴𝐸
And the relative axial displacement between C and d is 0:
ΔCd or Δ2 = 0
∆2 = ∑
𝑃(𝜕𝑃⁄𝜕𝑋2 )𝐿
=0
𝐴𝐸
Tabulation of bar forces:
Bar
L
P
aB
15
-46.875
0
BC
9
-18.75-0.6X1
CD
9
De
𝑷(𝝏𝑷⁄𝝏𝑿𝟏 )𝑳
𝑷(𝝏𝑷⁄𝝏𝑿𝟐 )𝑳
0
0
0
-0.6
0
101.25+3.24X1
0
-18.75-0.6X2
0
-0.6
0
101.25+3.24X2
15
-15.625
0
0
0
0
ab
9
28.125
0
0
0
0
bc
9
28.125-0.6X1
-0.6
0
-151.875+3.24X1
0
cd
9
9.375-0.6X2
0
-0.6
0
-50.625+3.24X2
de
9
9.375
0
0
0
0
Bb
12
50-0.8X1
-0.8
0
-480+7.68X1
0
Bc
15
-15.625+X1
1
0
-234.375+15X1
0
Cb
15
X1
1
0
15X1
0
Cc
12
-0.8X1-0.8X2
-0.8
-0.8
7.68X1+7.68X2
7.68X1+7.68X2
Cd
15
X2
0
1
0
15X2
Dc
15
15.625+X2
0
1
0
234.375+15X2
Dd
12
-0.8X2
0
-0.8
0
7.68X2
-765+51.84X1+7.68X2
285+7.68X1+51.84X2
𝝏𝑷⁄𝝏𝑿𝟏 𝝏𝑷⁄𝝏𝑿𝟐
sum
∆1 = ∑
𝑃(𝜕𝑃⁄𝜕𝑋1 )𝐿
=0
𝐴𝐸
-765+51.84X1+7.68X2 = 0
(Equation 1)
∆2 = ∑
𝑃(𝜕𝑃⁄𝜕𝑋2 )𝐿
=0
𝐴𝐸
285+7.68X1+51.84X2= 0
(Equation 2)
And solving for X1 and X2,
Note: Negative answer implies
𝑋
15.92
{ 1} = {
}
𝑋2
−7.86
X1 = 15.92 kN (C)
opposite of the assumed
direction. Thus, we remark the
correct direction in our final
X2 = 7.86 kN (T)
answer.
Where X1 is the actual bar force at Cb, and X2 is the actual bar force at Cd.
5. Solve for actual bar forces:
Bar
P
X1
X2
P* = P(X1, X2)
aB
-46.875
15.92
-7.86
-46.875
BC
-18.75-0.6X1
15.92
-7.86
-28.302
CD
-18.75-0.6X2
15.92
-7.86
-14.034
De
-15.625
15.92
-7.86
-15.625
ab
28.125
15.92
-7.86
28.125
bc
28.125-0.6X1
15.92
-7.86
18.573
cd
9.375-0.6X2
15.92
-7.86
14.091
de
9.375
15.92
-7.86
9.375
Bb
50-0.8X1
15.92
-7.86
37.264
Bc
-15.625+X1
15.92
-7.86
0.295
Cb
X1
15.92
-7.86
15.92
Cc
-0.8X1-0.8X2
15.92
-7.86
-6.448
Cd
X2
15.92
-7.86
-7.86
Dc
15.625+X2
15.92
-7.86
7.765
Dd
-0.8X2
15.92
-7.86
6.288
Actual Bar forces:
-6.448
37.264
Ha = 0
-14.034 D
6.288
a
C
-28.302
B
28.125 b
18.573
Va = 37.5
50
c
14.091
d
e
9.375
8m
8m
8m
8m
Ve = 12.5