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

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

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

Slope Real load Virtual unit couple C A B w C A B x1 x2 1 RA RB rA rB qC 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

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

Displacement at C Virtual Moment mD Real Moment M x1 x2 x1 x2 SOLUTION B 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

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

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

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 = = 0.00469 m = 4.69mm, Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

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 = 0.00125 rad, Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

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

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 = 0.00172 m = 1.72 mm, Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

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 = 0.000194 rad Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

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

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 = + 28.8 mm , Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

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 = + 0.00125 rad , Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

DCH= = 28.87 mm A B C 5 m 6 m 2 kN/m 4 kN 12 kN 16 kN qC = 0.00125 rad , + 60 M , kN•m 16 -12 - + V , kN 4 Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

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

Displacement at C Virtual Moment mD Real Moment M 3 m B C 30o 1 kN x1 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 = 11.25 mm , Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

Slope at C Virtual Moment mq Real Moment M 5 kN 1 kN•m 3 m B C 30o 3 m x1 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 = 0.00875 rad, Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

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

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

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

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

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

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, = 0.026 mm, Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan = 2 + 0.026 = 2.03 mm,

- 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 = -2.31 mm , Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

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

= + 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

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

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

Real load x2 A B C 5 m 6 m 2 kN/m 4 kN 4 4 B C Axial, N (kN) A 12 60 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

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 = 0.0111 mm, Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

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 = 0.048 mm, Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

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 = 0.0288 m = + 28.8 mm , Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan

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 = 0.0173 m = 17.3 mm , - Axial Department of Civil Engineering University of Engineering and Technology, Taxila, Pakistan DCH = 0.00105 m = 1.05 mm ,

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

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