Method of Virtual Work Frame Deflection Example Steven Vukazich San Jose State University

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1 Method of Virtual Work Frame Deflection xample Steven Vukazich San Jose State University

2 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

3 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.

4 Find the Horizontal Deflection at Point Virtual system to measure δ " D ft 5 ft ft

5 Find Support Reactions for the Virtual System D ft y & M % = + y =.34 x + & F - = x = y 5 ft ft + & F. = y =.34

6 Support Reactions for the Virtual System D ft ft ft

7 hoose Sign onvention for Internal moment For horizontal member D + For vertical member +

8 Moment Diagram for the Virtual System ft D ft 5 ft ft

9 Support Reactions for the Real System Results from our previous module: k D 2 k ft 8 k k k 5 ft ft

10 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

11 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

12 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

13 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 k-ft 3 2 k-ft M 3 M DM L + c k-ft 3

14 valuate Product Integrals M Q M P 3 M DM L L k-ft M M k-ft 3 M ft k-ft M 3 3 M DM L ft L 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.

15 valuate Product Integrals 5 δ " = I 3 M 4 M 7 dx Segment X k-ft 3 Segment XD k-ft 3 5 KRS 3 M 4 M 7 dx = k ft k ft 3 2 in 3 ft 3 = 3,8.3 k in 3 Segment k-ft 3 Segment 7.7 k-ft 3 5 JKL 3 M 4 M 7 dx = 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.

16 valuate Product Integrals 5 JKL 3 M 4 M 7 dx = k ft 3 2 in 3 ft 3 = 87,92 k in 3 5 KRS 3 M 4 M 7 dx = k ft 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

17 Find the the Rotation at Point Virtual system to measure θ % D ft 5 ft ft

18 Find Support Reactions for the Virtual System D ft x y y & M % = + + & F - = y =.99 x = 5 ft ft + & F. = y =.99

19 Support Reactions for the Virtual System D.99 ft.99 5 ft ft

20 Moment Diagram for the Virtual System ft D 5 ft ft

21 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

22 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

23 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

24 valuate Product Integrals L 3 ft X X 2 ft M Q M 2 M M M P 8 k-ft M 3 d L D ft c M D + 2M H M L k-ft 2 M 3 2 k-ft M DM L + c k-ft 2

25 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

26 valuate Product Integrals 5 δ " = I 3 M 4 M 7 dx Segment X k-ft 2 Segment XD k-ft 2 Segment 5. k-ft 2 Segment 5 KRS 3 M 4 M 7 dx = 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

27 5 KRS 3 M 4 valuate Product Integrals M 7 dx = 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 in 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

28 Frame Deflection xample Results.2 in k ft D 8 k.52 rad 5 ft ft

k 21 k 22 k 23 k 24 k 31 k 32 k 33 k 34 k 41 k 42 k 43 k 44

k 21 k 22 k 23 k 24 k 31 k 32 k 33 k 34 k 41 k 42 k 43 k 44 CE 6 ab Beam Analysis by the Direct Stiffness Method Beam Element Stiffness Matrix in ocal Coordinates Consider an inclined bending member of moment of inertia I and modulus of elasticity E subjected shear