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

ContentsMethod of Virtual Work: Beams Frames

Method of Virtual Work : Beams and Frames Real load Vertical Displacement

Slope Real load

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.

SOLUTIONVirtual Moment mD Displacement at CmD2 = -x2M2 = -Px2

Virtual Moment mq Real Moment MSlope at CM2 = -Px2

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.

SOLUTION-1x =Vertical Displacement at B= 0.00469 m = 4.69mm,-1x =

SOLUTION-1 =Slope at B= 0.00125 rad,-1 =

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.

Virtual Moment mD Real Moment M= 0.00172 m = 1.72 mm,M1 = 14 - x1M2 = 6x2Displacement at B

Virtual Moment mq Real Moment M= 0.000194 radM1 = 14 - x1M2 = 6x2mq1 = 0.25x1mq2 = -0.25x2Slope at B

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 GPaI = 200(106) mm4

Real Moment MM2= 12 x2m2= 1.2 x2m1= x1M1= 16 x1- x12

16 kN12 kNReal Moment MVirtual Moment mq = + 0.00125 rad , M2= 12 x2m2= 1-x2/5m1= 0M1= 16 x1- x12

Example 8-22

Determine the slope and the vertical displacement of point C on the frame.Take E = 200 GPa, I = 15(106) mm4.

Virtual Moment mD Displacement at CmD1 = -0.5x1mD1 = -0.5x1M1 = -2.5x1mD2 = -1.5mD2 = -1.5M2 = -7.5= 11.25 mm ,

Virtual Moment mqReal Moment Mmq1 = -1mq2 = -17.5 kNmM1 = -2.5x1M2 =- 7.5= 0.00875 rad, Slope at Cmq1 = -1M1 = -2.5x1mq2 = -1M2 = -7.5

Virtual Strain Energy Caused by Axial Load, Shear, Torsion, and Temperature Axial LoadWheren = internal virtual axial load caused by the external virtual unit loadN = internal axial force in the member caused by the real loadsL = length of a memberA = cross-sectional area of a memberE = modulus of elasticity for the material

TorsionWhere t = internal virtual torque caused by the external virtual unit loadT = internal torque in the member caused by the real loadsG = 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

Shear

TemperatureT1T2T2 > T1

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.

SOLUTION Part (a) :

Part (b) : Vertical displacement at CSOLUTION0.5x0.5xDCv = -2.31 mm , - Bending

Part (b) : Horizontal displacement at BDBH = 0.84 mm , - Axial

Part (c) :

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.

Virtual load

Real load

Due to Axial= 1.109(10-5) m = 0.0111 mm,

Due to Shear= 4.8(10-5) m = 0.048 mm,

Due to Bending

Due to Temperature - Bending- Axial

Total Displacement= 0.01109 + 0.048 + 28.8 + (17.3 + 1.05) = 47.21 mm