Axial Deformation
Axial Deformation
Axial Deformation
AXIAL DEFORMATION
Prepared by:
ENGR. RANDY G. POLICARPIO
WORKING STRESS and FACTOR OF SAFETY
SAINT-VENANTS PRINCIPLE
SAINT-VENANTS PRINCIPLE
SAINT-VENANTS PRINCIPLE
ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER
Bar having cross-sectional area that gradually varies along its length L
𝑃𝐿
𝛿=
𝐴𝐸
where:
𝛿 − 𝑡𝑜𝑡𝑎𝑙 𝑑𝑒𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛
𝑃 − 𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝑙𝑜𝑎𝑑
𝐿 − 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑏𝑎𝑟
𝐴 − 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎
𝐸 − 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑖𝑡𝑦
Example Problem: : Reference: Mechanics of Materials by R C Hibbeler
A-36 steel bar shown in 𝐹𝑖𝑔. 4 − 6𝑎 is made from two segments having cross-sectional area of
𝐴𝐴𝐵 = 1 𝑖𝑛2 and 𝐴𝐵𝐷 = 2 𝑖𝑛2 . Determine the vertical displacement of end 𝐴 and the displacement
of 𝐵 relative to 𝐶.
12 𝑖𝑛 12 𝑖𝑛 12 𝑖𝑛
+15 𝑘𝑖𝑝𝑠 2 𝑓𝑡 +7 𝑘𝑖𝑝𝑠 1.5 𝑓𝑡 −9 𝑘𝑖𝑝𝑠 1 𝑓𝑡
1 𝑓𝑡 1 𝑓𝑡 1 𝑓𝑡
𝛿𝐴 = 𝑘𝑖𝑝 + 𝑘𝑖𝑝 + 𝑘𝑖𝑝
1 𝑖𝑛2 29000 2 2 𝑖𝑛2 29000 2 2 𝑖𝑛2 29000 2
𝑖𝑛 𝑖𝑛 𝑖𝑛
𝛿𝐴 = +0.0127 𝑖𝑛.
Since the result is positive, the bar elongates and so the displacement at A is upward
b. Solve for the displacement at B relative to C
𝑃𝐵𝐶 𝐿𝐵𝐶
𝛿𝐵𝐶 =
𝐴𝐵𝐶 𝐸
12 𝑖𝑛
+7 𝑘𝑖𝑝𝑠 1.5 𝑓𝑡
1 𝑓𝑡
𝛿𝐵𝐶 = 𝑘𝑖𝑝
2 𝑖𝑛2 29000 2
𝑖𝑛
C B
Solution:
a. Solve for the internal forces in the aluminum rod and steel rod.
C B
C B
𝛿 = 1.78 + 𝑦 {
By ratio and proportion:
𝑦 1.94−1.78
=
3.5 6
3.5
𝑦= 0.16 𝑚𝑚 = 0.093 𝑚𝑚
6
𝛿 = 1.78 𝑚𝑚 + 0.093 𝑚𝑚
𝛿 = 1.873 𝑚𝑚
Example Problem: : Reference: Mechanics of Materials by R C Hibbeler
A member is made from a material that has a specific weight 𝛾 and modulus of elasticity 𝐸. If it is in
the form of a cone having the dimensions shown in 𝐹𝑖𝑔. 4 − 9𝑎., determine how far its end is
displaced due to gravity when it is suspended in the vertical position.
Solution:
𝑥 𝑟𝑜
=
𝑦 𝐿
𝑟𝑜
𝑥= 𝑦
𝐿
𝐹𝑖𝑔. 4 − 9
𝐹𝑟𝑜𝑚 𝐹𝑖𝑔. 4 − 9𝑏
𝑃 𝑦 − 𝑊(𝑦) = 0
𝑃(𝑦) − 𝛾𝑉 = 0