Theory of Structures
Elastic Deflections of Structures
Deflections of Frames Using Unit Load Method (Virtual Work)
Example: Determine the horizontal displacement at joint B and slope at joint A of the frame shown below.
Take E = 200 GPa, I = 500x106 mm4.
Solution:
3m
B
1 kN
C
B
C
F
B
F
3m
C
2I
I
I
D
D
1/6kN
1.25kN
A
A
I
48 kN
E
E
I
D
4.5m
E
5m
3m
F
96 kN
84 kN
A
1 kN
48 kN
1 kN.m
1.25kN
1/6kN
m,A
Segment Origin
AE
A
BE
B
BF
B
CF
C
CD
C
𝑙
𝑀𝑚𝑑𝑥
∆𝐵ℎ = ∫
𝐸𝐼
0
=∫
12 kN
m,B)h
Limit
0 4.5
0 3
0 3
0 3
0 5
M
mhB
48x
1x
216
7.5-1x
216+12x 7.5-1.25x
84x
1.25x
0
0
M
mA
1.0
1.0
1.0-x/6
x/6
0
4.5 (
3
3(
3(
48𝑥 )(𝑥 )𝑑𝑥
216(7.5 − 𝑥 )𝑑𝑥
216 + 12𝑥 )(7.5 − 1.25𝑥 )𝑑𝑥
84𝑥 )(1.25𝑥 )𝑑𝑥
+∫
+∫
+∫
𝐸𝐼
𝐸𝐼
𝐸𝐼
2𝐸𝐼
0
0
0
0
175 2
125 2
3 (−25𝑥 −
𝑥 − 175𝑥) 𝑑𝑥
𝑥 ) 𝑑𝑥
6
6
+∫
𝐸𝐼
𝐸𝐼
0
3(
1. ∆𝐵ℎ = ∫
0
1. ∆𝐵ℎ =
1
2
3 3
{[(16𝑥 3 )]4.5
0 + [3240𝑥 − 198𝑥 + 12.5𝑥 ]0 }
𝐸𝐼
1. ∆𝐵ℎ =
1458 + 9720 − 1782 + 337.5
9733.5
=
6
𝐸𝐼
200 ∗ 10 ∗ (500 ∗ 106 ) ∗ 10−12
∆𝐵ℎ = 97.33 𝑚𝑚 →
1. 𝜃𝐴 = ∫
ddownward
𝑥
𝑥
3 (216 + 12𝑥 ) (1.0 − ) 𝑑𝑥
3 (84𝑥 ) ( ) 𝑑𝑥
48𝑥 )(1.0)𝑑𝑥
216(1.0)𝑑𝑥
6
6
+∫
+∫
+∫
𝐸𝐼
𝐸𝐼
𝐸𝐼
2𝐸𝐼
0
0
0
4.5 (
0
3
𝜃𝐴 = 0.0146 𝑟𝑎𝑑 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
Dr. Nibras Nizar
ddownward
37
Theory of Structures
Elastic Deflections of Structures
Example: Determine the rotation on the left and right sides of internal hinge B for the frame shown below.
C
C
1/4
kN
D
2I
1 kN.m B
3m
10 kN
1 kN.m
B
C
20 kN
B
A
1/4
kN
1 kN.m
0 2.5
mBA
-1.0
0
0
1/4
10 kN
B
mBC
0
1-(1/4(4x/5))
=1-x/5
1/4(4x/5)
=x/5
1
B
B
m
M
m
Seg.AB
M
x
𝑙
𝑀𝑚𝑑𝑥
𝐸𝐼
0
1. 𝜃 = ∫
B
B
B
0
1/4
C
x
 3
4
M
𝑥
𝑥
) 𝑑𝑥 2.5 (8𝑥 ) ( ) 𝑑𝑥
5
5
∫
2𝐸𝐼
2𝐸𝐼
0
C
x
3

 3
1

m
C
 3
4
10
2.5 (8𝑥 ) (1 −
4
m

1/4
Seg.CD
2.5
8𝑥2
12.5
𝜃𝐵𝐶 = [
] =
2 ∗ 2𝐸𝐼 0
𝐸𝐼
𝜃𝐵𝐴 =
Dr. Nibras Nizar
18
𝑟𝑎𝑑 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
𝐸𝐼
 3
4
Seg.BD
2𝑥3
18
𝜃𝐵𝐴 = [
] =
3𝐸𝐼 0 𝐸𝐼
0
x
 3
4

−2𝑥 )(−1.0)𝑑𝑥
+0+0
𝐸𝐼
1. 𝜃𝐵𝐶 = 0 + ∫
x
 3
4
10
3(
0
m
x
1. 𝜃𝐵𝐴 = ∫
x
C
M
-4x(x/2)=-2x
10(4x/5)
=8x
10(4x/5)
=8x
2m
x
CD
Limit
0 3
0 2.5
A 12 kN
18
kN.m10 kN 2m
x
Segment Origin
AB
B
BD
B
I
4kN/m
A
3m
4kN/m
E
𝜃𝐵𝐶 =
12.5
𝑟𝑎𝑑 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒
𝐸𝐼
38
Theory of Structures
Elastic Deflections of Structures
Deflections of Trusses
The displacement of a truss joint due to an external loading, temperature change, or fabrication errors can be
determined by the method of virtual work using the following equation:
∆= ∑ 𝑢𝑑𝐿
𝑆𝐿
∆= ∑ 𝑢( + 𝛼𝐿∆𝑡 + ∆𝐿𝑓𝑎𝑏. )
𝐸𝐴
Where:
= Joint displacement caused by the real loads on the truss.
u= Virtual normal force in a truss member caused by the external virtual unit load.
S= Normal force in a truss member caused by the real loads.
L= Length of a member.
A= Cross sectional area of a member.
E= Modulus of elasticity of a member.
a= Coefficient of thermal expansion of member.
T= Change in temperature of member.
Lfab.= Difference in length of the member from its intended size as caused by a fabrication error.
Sign Convention
u, S =
+ if tension
- if compression
T = + if increase in temp.
- if decrease in temp.
Lfab. = + if too long
- if too short
Example: Determine the horizontal and vertical components of the deflection at joint B of the truss shown
below by virtual work method. EA constant, E=200GPA, A=1200mm2
C
1kN A
D
4m
B
A
D
1kN
3m
4m
virtual Truss (u,B)h)
B
4m
35kN A
3m
1kN
0.43kN
D
S
u,B)h
u,B)v
SLu,B)h
SLu,B)v
AB
BC
AD
BD
CD
4
3
5.66
4
5
21
21
-79.2
84
-35

1.0
0
0
0
0
0.43
0.43
-0.6
1.0
-0.71
84
0
0
0
0
84
36.12
27.09
273.45
336.0
124.25
796.91
∆𝐵)ℎ = 0.35 𝑚𝑚 →
3m
84kN
56kN
virtual Truss (u,B)v)
L
B
4m
0.57kN
Member
Dr. Nibras Nizar
C
4m
35kN
4m
C
28kN
Real Truss, S
∆𝐵)ℎ =
∆𝐵)ℎ =
∆𝐵)𝑣 =
∆𝐵)ℎ =
1
𝐸𝐴
∑ 𝑢, ∆B)h 𝑆𝐿
84
200𝑥106 𝑥1200𝑥10−6
1
𝐸𝐴
= 0.00035𝑚
∑ 𝑢, ∆B)v 𝑆𝐿
796.91
= 0.00332𝑚
200𝑥106 𝑥1200𝑥10−6
∆𝐵)𝑣 = 3.32 𝑚𝑚 ↓
39
Theory of Structures
Elastic Deflections of Structures
D
4m
3m
1/2kN
B
60kN A
3m
1kN
E
(300)
(800)
(300)
0)
(50
(800)
C
A
F
60kN
(300)
(800)
(300)
B
3m
C
3m
40kN
1/2kN
D
4m
E
F
(50
0)
Example: Determine the vertical deflection at joint B for the truss shown below.
Members AB, BC, BF, BD increase in Temp. by 50oC, Member BD is 1.0 cm too short.
=12x10-6, E=2x105Mpa, Numbers inside brackets indicate A(mm2).
40kN
virtual Truss (u,B)v)
Real Truss, S
∆= ∑ 𝑢𝑑𝐿
∆= ∑ 𝑢 (
𝑆𝐿
𝐸𝐴
Member
AB
BC
FE
ED
AF
BE
CD
BF
BD
+ 𝛼𝐿∆𝑡 + ∆𝐿𝑓𝑎𝑏. )
𝑳 𝒎𝒎
(
)
𝑬𝑨 𝑵
3000
300(2).105
𝑺𝑳
𝜶𝑳∆𝒕
(𝒎𝒎)
(mm)
𝑬𝑨
S (N)
=5*10-5
5*10-5
5*10-5
5*10-5
4000
=5*10-5
800(2).105
2.5*10-5
2.5*10-5
5000
=5*10-5
500(2).105
-5
5*10
∆𝑳𝒇𝒂𝒃.
(mm)
dL
(mm)
u (N)
udL
60*103
0
-30*103
-30*103
40*103
0
-40*103
3.0
0
-1.5
-1.5
+1.0
0
-1.0
+1.8
+1.8
0
0
0
0
0
0
0
0
0
0
0
0
+4.8
+1.8
-1.5
-1.5
+1.0
0
-1.0
0
0
-3/8
-3/8
-1/2
0
-1/2
0
0
+0.5625
+0.5625
-0.5
0
+0.5
-50*103
50*103
-2.5
+2.5
+3.0
+3.0
0
-10.0
+0.5
-4.5
+5/8
+5/8
+0.3125
-2.8125
B)vudL
-1.375
∆𝐵)𝑣 = 1.375 𝑚𝑚 ↑
Dr. Nibras Nizar
40