Method of Virtual Work Frame Deflection xample Steven Vukazich San Jose State University
Frame Deflection xample 9 k k D 4 ft θ " # The statically determinate frame from our previous internal force diagram example is made up of columns that are W8x48 (I = 84 in 4 ) members and a beam that is a Wx22 (I = 8 in 4 ). The modulus of elasticity of structural steel is 29, ksi, which yields the following bending stiffnesses: I = 5,33, k-in 2 I D = 3,422, k-in 2 For the loads shown, find the following:
Frame Deflection xample ft k δ " D θ % 8 k The statically determinate frame from our previous internal force diagram example is made up of columns that are W8x48 (I = 84 in 4 ) members and a beam that is a Wx22 (I = 8 in 4 ). The modulus of elasticity of structural steel is 29, ksi, which yields the following bending stiffenesses: I = 5,33, k-in 2 I D = 3,422, k-in 2 For the loads shown, find the following: 5 ft ft. The horizontal deflection at point ; 2. The slope at the pin support at.
Find the Horizontal Deflection at Point Virtual system to measure δ " D ft 5 ft ft
Find Support Reactions for the Virtual System D ft y & M % = + y =.34 x + & F - = x = y 5 ft ft + & F. = y =.34
Support Reactions for the Virtual System D ft.34.34 5 ft ft
hoose Sign onvention for Internal moment For horizontal member D + For vertical member +
Moment Diagram for the Virtual System ft D ft 5 ft ft
Support Reactions for the Real System Results from our previous module: k D 2 k ft 8 k k k 5 ft ft
Moment Diagram for the Real System Results from our previous module: 8 k-ft 8 k-ft ft k-ft D 2 k-ft 5 ft ft
valuate the Virtual Work Product Integrals ft M Q 8.3x = 3.9 ft ft D X ft M P 8 k-ft 8 k-ft k-ft D 2 k-ft x X I = 5,33, k-in 2 I D = 3,422, k-in 2 5 ft ft δ " = I 3 M 4 M 7 dx 5 5 ft ft 8 x = 2 5 x Use Table to evaluate product integrals 9 = 3x x = 3 ft
Table to valuate Virtual Work Product Integrals ppendix Table.2 and X Table is as useful tool to evaluate product integrals of the form: 5 3 M 4 M 7 dx XD
valuate Product Integrals M Q M 2 8 k-ft M 3 L 3 ft X M 3.9 ft X 2 ft 3.9 ft M d L D ft c M P M D + 2M H M L 3.9 + 2 8 8 3 44.89 k-ft 3 2 k-ft M 3 M DM L + c 3.9 2 8 + 3.548 k-ft 3
valuate Product Integrals M Q M P 3 M DM L L 3 8 8 8 8 k-ft M M 3 7.7 k-ft 3 M ft k-ft M 3 3 M DM L ft L 3 333.333 k-ft 3
valuate Product Integrals 5 δ " = I 3 M 4 M 7 dx Segment X 44.89 k-ft 3 Segment XD 3.548 k-ft 3 5 KRS 3 M 4 M 7 dx = 44.89 k ft 3 3.548 k ft 3 2 in 3 ft 3 = 3,8.3 k in 3 Segment 333.333 k-ft 3 Segment 7.7 k-ft 3 5 JKL 3 M 4 M 7 dx = 333.333 + 7.7 k ft 3 2 in 3 ft 3 = 87,92 k in 3
valuate Product Integrals 5 JKL 3 M 4 M 7 dx = 333.333 + 7.7 k ft 3 2 in 3 ft 3 = 87,92 k in 3 5 KRS 3 M 4 M 7 dx = 44.89 k ft 3 3.548 k ft 3 2 in 3 ft 3 = 3,8.3 k in 3 5 JKL δ " = 3 M I 4 %U" 5 KRS M 7 dx + 3 M I 4 "VW M 7 dx δ " = 87,92 k in3 3,8.3 k in3 + 5,33, k in2 3,422, k in 2 δ " =.3 in +.38 in =.2 in δ " =.2 inches to the right Positive result, so deflection is in the same direction as the virtual unit load
Find the the Rotation at Point Virtual system to measure θ % D ft 5 ft ft
Find Support Reactions for the Virtual System D ft x y y & M % = + + & F - = y =.99 x = 5 ft ft + & F. = y =.99
Support Reactions for the Virtual System D.99 ft.99 5 ft ft
Moment Diagram for the Virtual System ft D 5 ft ft
Moment Diagram for the Real System Results from our previous module: 8 k-ft 8 k-ft ft k-ft D 2 k-ft 5 ft ft
valuate the Virtual Work Product Integrals ft M Q +.99(3) =.7273 D X ft M P 8 k-ft 8 k-ft k-ft D 2 k-ft 3 ft X 3 ft 5 ft ft 5 ft ft I = 5,33, k-in 2 I D = 3,422, k-in 2 δ " = I 3 M 4 M 7 dx 5 Use Table to evaluate product integrals
Table to valuate Virtual Work Product Integrals ppendix Table.2 X Table is as useful tool to evaluate product integrals of the form: 5 3 M 4 M 7 dx XD
valuate Product Integrals L 3 ft X X 2 ft M Q M 2 M M.7273.7273 M P 8 k-ft M 3 d L D ft c M D + 2M H M L.7273 + 2 8 3 24.54545 k-ft 2 M 3 2 k-ft M DM L + c.7273 2 8 + 2.334 k-ft 2
valuate Product Integrals M Q M P 8 k-ft M k-ft M 3 2 M DM L ft L 2 5. k-ft 2
valuate Product Integrals 5 δ " = I 3 M 4 M 7 dx Segment X 24.54545 k-ft 2 Segment XD 2.334 k-ft 2 Segment 5. k-ft 2 Segment 5 KRS 3 M 4 M 7 dx = 24.54545 + 2.334 k ft 2 2 H in 2 ft 2 = 2.88 k in 2 5 JKL 3 M 4 M 7 dx = 5 + k ft 2 2 H in 2 ft 2 = 7,2 k in 2
5 KRS 3 M 4 valuate Product Integrals M 7 dx = 24.54545 + 2.334 k ft 2 2 H in 2 ft 2 = 2.88 k in 2 5 JKL 3 M 4 M 7 dx = 5 + k ft 2 2 H in 2 ft 2 = 7,2 k in 2 5 JKL θ % = 3 M I 4 %U" 5 KRS M 7 dx + 3 M I 4 UVW M 7 dx θ % = 7,2 k in2 2.88 k in2 + 5,33, k in2 3,422, k in 2 θ % =.349 rad.7 rad =.52 rad θ % =.52 radians clockwise Negative result, so rotation is in the opposite direction of the virtual unit moment
Frame Deflection xample Results.2 in k ft D 8 k.52 rad 5 ft ft