Deflection: Virtual Work Method; Trusses Theory of Structure - I

Contents External Work and Strain Energy Principle of Work and Energy Principle of Virtual Work Method of Virtual Work: Trusses

External Work and Strain Energy Most energy methods are based on the conservation of energy principle, which states that the work done by all the external forces acting on a structure, Ue, is transformed into internal work or strain energy, Ui. Ue = Ui External Work-Force. x F L D P Ue Eigen work D As the magnitude of F is gradually increased from zero to some limiting value F = P, the final elongation of the bar becomes D. F Eigen work

x F D´ F ´ + P L P L F´ Eigen work D D´ D P Displacement work (Ue)Total = (Eigen Work)P + (Eigen Work)F´ + (Displacement work) P

x (m) F 20 kN L L 1 cm 0.01 m 20 kN

x (m) F 0.01 20 kN L L 15 kN L 15 kN 0.75 cm Eigen work 0.75 cm 0.25 cm 15 kN 0.0075 Displacement work 5 kN

External Work-Moment. q M q´ M ´ + M Eigen work dq M q M Displacement work -----(8-12) Eigen work -----(8-13) -----(8-14)

Strain Energy-Axial Force. D N

Strain Energy-Bending x dx w P L M dx dq s e

Strain Energy-Torsion dx T c dq g t g J For reference: Strength of Material by Singer, Fourth Edition, Page 67-68

Strain Energy-Shear V g dx dy t g For reference: Strength of Material by Singer, Fourth Edition, Page 161-163

Principle of Work and Energy x -PL M diagram P x V M + SMx = 0:

Principle of Virtual Work Apply virtual load P´ first u L A P´ = 1 Virtual loadings 1 • D = Su • dL Then apply real load P1. A u L Real displacements D In a similar manner, dL Virtual loadings 1 • q = Suq • dL P1 Real displacements

P1 P2 B n2 n1 n3 n4 n5 n6 n7 n8 n9 N2 N1 N3 N4 N5 N6 N7 N8 N9 B 1kN Method of Virtual Work : Truss P1 P2 B External Loading. n2 n1 n3 n4 n5 n6 n7 n8 n9 N2 N1 N3 N4 N5 N6 N7 N8 N9 B 1kN D Where: 1 = external virtual unit load acting on the truss joint in the stated direction of D n = internal virtual normal force in a truss member caused by the external virtual unit load D = external joint displacement caused by the real load on the truss N = internal 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

Temperature Where: D = external joint displacement caused by the temperature change a = coefficient of thermal expansion of member DT = change in temperature of member Fabrication Errors and Camber Where: D = external joint displacement caused by the fabrication errors DL = difference in length of the member from its intended size as caused by a fabrication error

Example 8-15 The cross-sectional area of each member of the truss shown in the figure is A = 400 mm2 and E = 200 GPa. (a) Determine the vertical displacement of joint C if a 4-kN force is applied to the truss at C. (b) If no loads act on the truss, what would be the vertical displacement of joint C if member AB were 5 mm too short? (c) If 4 kN force and fabrication error are both accounted, what would be the vertical displacement of joint C. A B C 4 m 4 kN 3 m

SOLUTION Part (a) Virtual Force n. Since the vertical displacement of joint C is to be determined, only a vertical 1 kN load is placed at joint C. The n force in each member is calculated using the method of joint. Real Force N. The N force in each member is calculated using the method of joint. 1 kN A B C 4 kN N(kN) A B C n (kN) 0.667 -0.833 2 +2.5 -2.5 1.5 kN 4 kN 0.5 kN

= 1 kN C 4 kN -0.833 -0.833 C -2.5 +2.5 A A 0.667 B B 2 C n (kN) L (m) 5 5 = A B C nNL (kN2•m) 8 -10.41 10.41 10.67 DCv = 0.133 mm,

Part (b): The member AB were 5 mm too short C n (kN) 1 kN 0.667 -0.833 5 mm DCv = -3.33 mm, Part (c): The 4 kN force and fabrication error are both accounted. DCv = 0.133 - 3.33 = -3.20 mm DCv = -3.20 mm,

Example 8-16 Determine the vertical displacement of joint C of the steel truss shown. The cross-section area of each member is A = 400 mm2 and E = 200 GPa. 4 m A B C D E F 4 kN

SOLUTION Virtual Force n. Since the vertical displacement of joint C is to be determined, only a vertical 1 kN load is placed at joint C. The n force in each member is calculated using the method of joint. Real Force N. The N force in each member is calculated using the method of joint. 4 m A B C D E F n (kN) 0.667 -0.471 -0.943 0.333 1 -0.333 4 m A B C D E F 4 kN N(kN) 4 -5.66 -4 0.667 kN 0.333 kN 4 kN 1 kN

= A B C D E F n (kN) 1 kN A B C D E F 4 kN N(kN) A B C D E F L(m) A B 0.667 -0.471 -0.943 0.333 1 -0.333 A B C D E F n (kN) 1 kN 4 -5.66 -4 A B C D E F 4 kN N(kN) 4 5.66 A B C D E F L(m) A B C D E F nNL(kN2•m) = 10.67 15.07 30.18 5.33 16 DCv = 1.23 mm,

Example 8-17 Determine the vertical displacement of joint C of the steel truss shown. Due to radiant heating from the wall, members are subjected to a temperature change: member AD is increase +60oC, member DC is increase +40oC and member AC is decrease -20oC.Also member DC is fabricated 2 mm too short and member AC 3 mm too long. Take a = 12(10-6) , the cross-section area of each member is A = 400 mm2 and E = 200 GPa. 2 m A B C D 3 m 20 kN 10 kN wall

SOLUTION Due to loading forces. 0.667 kN 1 kN 1 kN 13.33 kN 23.33 kN 2 m A B C D 3 m n (kN) 2 m A B C D 3 m 20 kN 10 kN N (kN) 0.667 -1.2 1 23.33 -24.04 20 2 3.61 3 A B C D L (m) 31.13 104.12 60 A B C D nNL(kN2•m) DCv= 2.44 mm,

Fabrication error (mm) A B D C 1 kN 0.667 -1.2 1 A B C D n (kN) +40 -20 +60 A B D DT (oC) C 2 3.61 3 A B C D L (m) Fabrication error (mm) -2 + 3 A B D C Due to temperature change. = 3.84 mm, Due to fabrication error. = -4.93 mm, Total displacement . = 1.35 mm,

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