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