1. Deflection: Virtual Work Method; Beams and Frames
Theory of Structure - I

2. Contents Method of Virtual Work: Beams Frames
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

3. Method of Virtual Work : Beams and Frames
Vertical Displacement
Real load
Virtual unit load C A B w C A B x1 x2 RA
DCv RB rA 1 rB x1 x2 w B x2 RB V2 M2 x1 rA
vD1
mD1 B rB x2
vD2
mD2 x1 RA V1 M1
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

4. Slope Real load Virtual unit couple C A B w C A B x1 x2 1 RA RB rA rBqC x1 x2 w B x2 RB V2 M2 B rB x2
vq2
mq2 x1 RA V1 M1 rA
vq1
mq1
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

5. Example 8-18
The beam shown is subjected to a load P at its end. Determine the slope and displacement at C. EI is constant. 2a a A B C P nC
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

6. Displacement at C Virtual Moment mD Real Moment M x1 x2 x1 x2 SOLUTIONB C 2a a P
1 kN x1 x2 x1 x2 A B C 2a a M m -a
-Pa
mD2 = -x2
mD2 = -x2
M2 = -Px2
M2 = -Px2
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

7. Slope at C Virtual Moment mq Real Moment M x1 x2 x1 x2 A B C 2a a P
1 kN•m x1 x2 A B C 2a a m M -1
-Pa
M2 = -Px2
M2 = -Px2
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

8. Example 8-19
Determine the slope and displacement of point B of the steel beam shown in the figure below. Take E = 200 GPa, I = 250(106) mm4. A
5 m B
3 kN/m
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

9. Vertical Displacement at B
SOLUTION
Vertical Displacement at B
Virtual Moment mD A
5 m B
Real Moment M A
5 m B
3 kN/m x
1 kN x x 3x V M
1 kN x v mD
-1x =
-1x =
= m = 4.69mm,
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

10. Slope at B Virtual Moment mq Real Moment M SOLUTION A 5 m B A 5 m B
3 kN/m x x
1 kN•m x 3x V M
-1 =
-1 = x v mq
1 kN•m
= rad,
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

11. Example 8-20
Determine the slope and displacement of point B of the steel beam shown in the figure below. Take E = 200 GPa, I = 60(106) mm4. A C D B
5 kN
14 kN•m
2 m
3 m
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

12. Virtual Moment mD Displacement at B Real Moment M 1 kN x1 x2 x3 A C D
14 kN•m
2 m
3 m x1 x2 x3
0.5 kN
1 kN
6 kN 1 mD
M1 = 14 - x1
M1 = 14 - x1 14
M2 = 6x2
M2 = 6x2 mD
= m = 1.72 mm,
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

13. Slope at B Virtual Moment mq Real Moment M x1 x2 x3 A C D B 2 m 3 m
1 kN•m A C D B
5 kN
14 kN•m
2 m
3 m
0.25 kN
0.25 kN x1 x2 x3
1 kN
6 kN
mq1 = 0.25x1
mq1 = 0.25x1 mq
0.5
-0.5
M1 = 14 - x1
M1 = 14 - x1
mq2 = -0.25x2
mq2 = -0.25x2 14
M2 = 6x2
M2 = 6x2 mD
= rad
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

14. Example 8-21
(a) Determine the slope and the horizontal displacement of point C on the frame. (b) Draw the bending moment diagram and deflected curve.
E = 200 GPa
I = 200(106) mm4 A B C
5 m
6 m
2 kN/m
4 kN
1.5 EI EI
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

15. Real Moment M Virtual Moment mD A C x2 A B C 5 m 6 m 2 kN/m 4 kN x2
M2= 12 x2
m2= 1.2 x2 1
12 kN
1.2 kN
M2= 12 x2
m2= 1.2 x2
M1= 16 x1- x12
M1= 16 x1- x12 x1
m1= x1
m1= x1 x1
16 kN
1 kN
12 kN
1.2 kN
= mm ,
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

16. Real Moment M Virtual Moment mq x2 A B C 5 m 6 m 2 kN/m 4 kN x2 1 kN•m
M2= 12 x2
m2= 1-x2/5
12 kN
1/5 kN
M2= 12 x2
m2= 1-x2/5
M1= 16 x1- x12
M1= 16 x1- x12 x1
m1= 0
m1= 0 x1
16 kN
1/5 kN
12 kN
= rad ,
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

17. DCH= = 28.87 mm A B C 5 m 6 m 2 kN/m 4 kN 12 kN 16 kN
qC = rad , + 60
M , kN•m 16
-12 - +
V , kN 4
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

18. Example 8-22
Determine the slope and the vertical displacement of point C on the frame.
Take E = 200 GPa, I = 15(106) mm4.
5 kN
3 m
60o
2 m A B C
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

19. Displacement at C Virtual Moment mD Real Moment M 3 m B C 30o 1 kNx1
1 kN C
30o
nD1
vD1 x1
5 kN C
30o N1 V1 x1 x1
1.5 m
1.5 m
mD1 = -0.5x1
mD1 = -0.5x1
5 kN
M1 = -2.5x1
1 kN
7.5 kN•m
1.5 kN•m x2
1.5 kN•m
1 kN
vD2
nD2 x2
7.5 kN•m
5 kN N2 V2
2 m A
1.5 kN•m
2 m A
7.5 kN•m x2 x2
mD2 = -1.5
mD2 = -1.5
M2 = -7.5
= mm ,
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

20. Slope at C Virtual Moment mq Real Moment M 5 kN 1 kN•m 3 m B C 30o 3 mx1 C
30o
nq1
vq1
1 kN•m x1
5 kN C
30o N1 V1 x1 x1
1.5 m
1.5 m
mq1 = -1
mq1 = -1
M1 = -2.5x1
M1 = -2.5x1
1 kN•m
7.5 kN•m
5 kN x2
7.5 kN•m
5 kN N2 V2
2 m A
1 kN•m x2
1 kN•m
nq2
vq2
7.5 kN•m x2
2 m A x2
mq2 = -1
mq2 = -1
M2 = -7.5
M2 =- 7.5
= rad,
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

21. N = internal axial force in the member caused by the real loads
Virtual Strain Energy Caused by Axial Load, Shear, Torsion, and Temperature
Axial Load
Where
n = internal virtual axial load caused by the external virtual unit load
N = internal axial force in the member caused by the real loads
L = length of a member
A = cross-sectional area of a member
E = modulus of elasticity for the material
Bending
Where
n = internal virtual moment cased by the external virtual unit load
M = internal moment in the member caused by the real loads
L = length of a member
E = modulus of elasticity for the material
I = moment of inertia of cross-sectional area, computed about the the neutral axis
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

22. t = internal virtual torque caused by the external virtual unit load
Torsion
Where
t = internal virtual torque caused by the external virtual unit load
T = internal torque in the member caused by the real loads
G = shear modulus of elasticity for the material
J = polar moment of inertia for the cross section, J = pc4/2, where c is the
radius of the cross-sectional area
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

23. by the external virtual unit load
Shear
Where
v = internal virtual shear in the member, expressed as a function of x and caused
by the external virtual unit load
V = internal shear in the member expressed as a function of x and caused by the
real loads
K = form factor for the cross-sectional area:
K = 1.2 for rectangular cross sections
K = 10/9 for circular cross sections
K 1 for wide-flange and I-beams, where A is the area of the web
G = shear modulus of elasticity for the material
A = cross-sectional area of a member
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

24. Temperature T2 > T1 T1 dx T2 T2 > T1 T1 O dq DT = T2 - T1 T2 cy dq M y
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

25. From the beam below Determine :
Example 8-23
From the beam below Determine :
(a) If P = 60 kN is applied at the midspane C, what would be the displacement at point C. Due to shear and bending moment.
(b) If the temperature at the top surface of the beam is 55 oC , the temperature at the bottom surface is 30 oC and the room temperature is 25 oC.
What would be the vertical displacement of the beam at its midpoint C and the the horizontal deflection of the beam at support B.
(c) if (a) and (b) are both accounted, what would be the vertical displacement of the beam at its midpoint C.
Take a = 12(10-6)/oC. E = 200 GPa, G = 80 GPa, I = 200(106) mm4 and A = 35(103) mm2. The cross-section area is rectangular. A B C
2 m
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

26. Part (a) : SOLUTION A B C 2 m P A B 1 kN x x x P/2 0.5 kN P/2 V
diagram
0.5 v
diagram M
diagram
PL/4 m
diagram
0.5x 1
= 2 mm,
= mm,
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan
= = 2.03 mm,

27. - Bending SOLUTION Part (b) : Vertical displacement at C A B C 2 m
55 oC,
30 oC
260 m
Troom = 25 oC ,
T1=55oC
T2=30oC
260 mm
Temperature profile
1 kN x x A B
0.5 kN m
diagram 1
0.5x
0.5x
- Bending
DCv = mm ,
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

28. Part (b) : Horizontal displacement at B
55 oC
30 oC
260 m
Troom = 25 oC ,
T1=55oC
T2=30oC
260 mm
Temperature profile x
1 kN A B
1 kN n
diagram 1 1
- Axial
DBH = 0.84 mm
DCv = 2.31 mm ,
Deflected curve A B C
DBH = 0.84 mm ,
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

29. = + Part (c) : P A B C 55 oC 30 oC 260 m A B C DC = 2.03 mm P
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

30. Example 8-24
Determine the horizontal displacement of point C on the frame.If the temperature at top surface of member BC is 30 oC , the temperature at the bottom surface is 55 oC and the room temperature is 25 oC.Take a = 12(10-6)/oC, E = 200 GPa, G = 80 GPa, I = 200(106) mm4 and A = 35(103) mm2 for both members. The cross-section area is rectangular. Include the internal strain energy due to axial load and shear. A B C
5 m
6 m
2 kN/m
4 kN
1.5 EI,1.5AE, 1.5GA
EI,AE,GA
260 mm
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

31. Virtual load x2 5 m 6 m A C B 1 1 1 B C Axial, n (kN) A 1.2 kN 1.2 x1
Shear, v (kN) A 6
1.2x2 1 B C
Moment, m (kN•m) A
-1.2 6
-1.2
1x1 1
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

32. Real load x2 A B C 5 m 6 m 2 kN/m 4 kN 4 4 B C Axial, N (kN) A 1260
12x2 B C
Shear, V (kN) A 4 B C
Moment, M (kN•m) A 60
-12
16 - 2x1
-12
16x1 - x12 16
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

33. Due to Axial x2 B C Virtual Axial, n (kN) A 1.2 1 B C
Real Axial, N (kN) A 12 4 AE
6 m
1.5AE x1
5 m
= 1.109(10-5) m = mm,
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

34. Due to Shear x2 GA B C Virtual Shear, v (kN) A 1 -1.2 B C
Real Shear, V (kN) A 16 4
16 - 2x1
-12
6 m
1.5GA x1
5 m
= 4.8(10-5) m = mm,
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

35. Due to Bending x2 B C Real Moment, M (kN•m) A 60 16x1 - x12 12x2 B C
Virtual Moment, m (kN•m) A 6
1x1
1.2x2 EI
6 m
1.5EI x1
5 m
= m = mm ,
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

36. Due to Temperature - Bending - Axial B C m (kN•m) A 6 1x1 1.2x2 A B C
260 mm
30oC
55oC
Troom = 25oC B C
n (kN) A
1.2 1
- Bending
T1=30oC
T2=55oC
260 mm
Temperature
profile
Tm= 42.5oC
DCH = m = mm ,
- Axial
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan
DCH = m = mm ,

37. Total Displacement = 0.01109 + 0.048 + 28.8 + (17.3 + 1.05) = 47.21 mm
DCH= mm A B C
2 kN/m
4 kN
Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan

38. Thank You Department of Civil Engineering
University of Engineering and Technology, Taxila, Pakistan