Vukazich Fall 2016 CE 160 Notes: Virtual Work Frame Example 1 k C 8 ft B D E 10 ft 8 k A 5 ft 6 ft The statically determinate frame from our internal force diagram example is made up of columns that are W8x48 (I = 184 in4) members and a beam that is a W10x22 (I = 118 in4). The modulus of elasticity of structural steel is 29,000 ksi, which yields the following bending stiffnesses: 2 • EIABC = 5,336,000 k-in 2 • EIBDE = 3,422,000 k-in Find the horizontal displacement of point C using the method of virtual work. Real Problem We found the Shear, Moment, and Axial force diagrams for this frame in a previous example earlier in the semester. 1 CE 160 Virtual Work Frame Example Vukazich Fall 2016 Moment Diagram of Real or P-System (MP diagram) (You should review your notes and verify this is the correct diagram) C 18 k-ft 8 k-ft 10 k-ft D E B 12 k-ft 6 ft A 8 ft Virtual Problem Virtual System to measure δC 1 C 8 ft B D E 10 ft A 5 ft 6 ft 2 CE 160 Virtual Work Frame Example Vukazich Fall 2016 Free Body Diagram of Virtual System (or Q-system) 1 C 8 ft Ey B D 10 ft Ax Ay 5 ft 6 ft Equilibrium �! = 0; �! = 0; �! = 0 yields the following support reactions: 1 C 8 ft B D E 1.6364 10 ft A 1 1.6364 5 ft 6 ft 3 CE 160 Virtual Work Frame Example Vukazich Fall 2016 Virtual Moment Diagram (MQ diagram) (You should be able to verify this is the correct diagram) C 18 ft 8 ft 13.091 ft 10 ft E B D 8 ft A 4 CE 160 Virtual Work Frame Example Vukazich Fall 2016 Use Table 4 in the text to evaluate the virtual work product integrals that come from the principle of virtual work: 1 ! 1 ∙ �! = �!�! �� �� ! Integrate over three segments: AB, BC, and BDE MQ Diagram MP Diagram C C 18 k-ft 18 ft 8 ft 13.091 ft 8 k-ft E 10 ft E 10 k-ft D B D B 12 k-ft 6 ft 8 ft A 8 ft A Segment AB From Virtual Work Integral Table (see page 8): 1 1 � � � = 10 ft 10 k-ft 10 ft 3 ! ! 3 = 333.333 k-ft3 = 576,000 k-in3 Segment BC From Virtual Work Integral Table (see page 8): 1 1 � � � = 8 ft 8 k-ft 8 ft 3 ! ! 3 =170.6667 k-ft3 = 294,912 k-in3 5 CE 160 Virtual Work Frame Example Vukazich Fall 2016 Segment BDE -- Split into two segments at inflection point of MP diagram (point Z) 18 �� 12 �� = � 5 − � x = 3 ft MQ Diagram MP Diagram 13.091 ft 18 k-ft 18 ft E D E B Z B Z 12 k-ft 6 ft 8 ft 3 ft 3 ft 8 ft Segment BZ 18 ft 18 k-ft 13.091 ft B Z B Z 3 ft 3 ft From Virtual Work Integral Table (see page 8): 1 1 � + 2� � � = 13.091 ft +2 18 ft 18 k-ft 3 ft 6 ! ! ! 6 = 441.818 k-ft3 = 763,461.8167 k-in3 6 CE 160 Virtual Work Frame Example Vukazich Fall 2016 7 CE 160 Virtual Work Frame Example Vukazich Fall 2016 Segment ZDE 13.091 ft Z D E Z E 12 k-ft 8 ft 2 ft 6 ft From Virtual Work Integral Table (see page 8): Note that result is negative due to dissimilar curvatures 1 1 − � � � + � = − 13.091 ft 12 k-ft 8 ft + 6 ft 6 ! ! 6 = -366.5454 k-ft3 = -633,390.54 k-in3 Total for segment BDE 763,461.8167 k-in3 - 633,390.54 k-in3 = 130,071.28 Divide by EI of each segment 576,000 k-in3 294,912 k-in3 130,071.28 k-in3 � = + + ! 5,336,000 k-in2 5,336,000 k-in2 3,422,000 k-in2 �! = 0.1079 in + 0.05527 in + 0.03801 in �� = �. ��� in Result is positive, so deflection is in the direction of the unit load – to the right 8 CE 160 Virtual Work Frame Example Vukazich Fall 2016 Product Integral Evaluation using Table 4 in text: 9 CE 160 Virtual Work Frame Example .