13 160 virtual work truss example - powering silicon valley 160 virtual work truss... · 2 vukazich...
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1 VukazichCE160TrussDeflectionsusingMethodofVirtualWork[13]
CE 160 Notes: Truss Deflections Using Method of Virtual Work
A truss made of steel members is pin supported at point a, roller supported at point b and subjected to a point loads at points c and e as shown. The cross sectional areas of each member are shown on the figure. Find the horizontal and vertical deflection of point e using the Method of Virtual Work.
Use E = 30,000 ksi.
e
20 ft
15 ft 15 ft
c d
a b
24 k
20 ft 3 in2 3 in2
3 in2
1 in2
3 in2
1 in2
1 in2
12 k
δH#
δV#
2 VukazichCE160TrussDeflectionsusingMethodofVirtualWork[13]
Real System
Find truss member forces using method of joints (or method of sections)
Note: Tensile forces are positive
Virtual System to measure horizontal displacement at point e - δH
e
20 ft
30 ft
cd
a b
24 k
20 ft (20 k) (-16 k)
(-16 k)
(0)
(-20 k)
(40 k)
(12 k)32 k
36 k
32 k
12 k
3 VukazichCE160TrussDeflectionsusingMethodofVirtualWork[13]
Find truss member forces using method of joints (or method of sections)
Note: Tensile forces are positive
20 ft
1
20 ft
e
15 ft 15 ft
cd
a b
1e
20 ft
30 ft
cd
a b
20 ft (1.667) (-1.333)
(-1.333)
(0)
(1.667)
(0)
1 k
1.333 1.333
(0)
4 VukazichCE160TrussDeflectionsusingMethodofVirtualWork[13]
Method of Virtual Work to find δH
1 ∙ 𝛿! = 𝐹!"𝐹!"𝐿!𝐴!𝐸!
!
!!!
1 ∙ 𝛿! = 𝐹!"#𝐹!"#𝐿!"𝐴!"𝐸
+ 𝐹!"#𝐹!"#𝐿!"𝐴!"𝐸
+ 𝐹!"#𝐹!"#𝐿!"𝐴!"𝐸
+ 𝐹!"#𝐹!"#𝐿!"𝐴!"𝐸
1 ∙ 𝛿! = 1.66740 𝑘 25 𝑓𝑡 12 𝑖𝑛/𝑓𝑡3 𝑖𝑛! 30,000 𝑘𝑠𝑖 + 1.667
20 𝑘 25 𝑓𝑡 12 𝑖𝑛/𝑓𝑡3 𝑖𝑛! 30,000 𝑘𝑠𝑖
+ −1.333−16 𝑘 20 𝑓𝑡 12 𝑖𝑛/𝑓𝑡
3 𝑖𝑛! 30,000 𝑘𝑠𝑖 + −1.333−16 𝑘 20 𝑓𝑡 12 𝑖𝑛/𝑓𝑡
3 𝑖𝑛! 30,000 𝑘𝑠𝑖
𝛿! = 0.2222 𝑖𝑛 + 0.1111 𝑖𝑛 + 0.05689 𝑖𝑛 + 0.0569 𝑖𝑛
𝜹𝑯 = 𝟎.𝟒𝟒𝟕 𝒊𝒏 (positive, so deflection is in same direction as virtual load - to the right)
Virtual System to measure vertical displacement at point e – δV
Find truss member forces using method of joints (or method of sections)
20 ft
1
20 ft
e
15 ft 15 ft
cd
a b
5 VukazichCE160TrussDeflectionsusingMethodofVirtualWork[13]
Note: Tensile forces are positive
Method of Virtual Work to find δV
1 ∙ 𝛿! = 𝐹!"𝐹!"𝐿!𝐴!𝐸!
!
!!!
1 ∙ 𝛿! = 𝐹!"#𝐹!"#𝐿!"𝐴!"𝐸
+ 𝐹!"#𝐹!"#𝐿!"𝐴!"𝐸
1 ∙ 𝛿! = −1.0−16 𝑘 20 𝑓𝑡 12 𝑖𝑛/𝑓𝑡
3 𝑖𝑛! 30,000 𝑘𝑠𝑖 + −1.0−16 𝑘 20 𝑓𝑡 12 𝑖𝑛/𝑓𝑡
3 𝑖𝑛! 30,000 𝑘𝑠𝑖
𝛿! = 0.04267 𝑖𝑛 + 0.04267 𝑖𝑛
𝜹𝑽 = 𝟎.𝟎𝟖𝟓𝟑 𝒊𝒏 (positive, so deflection is in same direction as virtual load - down)
1
e
20 ft
30 ft
cd
a b
20 ft (0) (-1.0)
(-1.0)
(0)
(0)
(0)1.0
(0)