A Ignore Internal Strain Energy due toaxial load and shear.VA (pinned)15ff2 k/ffV V VC (hinge)15 ftNO€ (pireed)1(A) = St20 ftEI= constantE = 10,000 ksiI = 2,500 inM₂M- dxΕΙFind Horizontal DeflectionEI=ConstantE = 10,000 ks:I=2,500 in 4On 6/9at joint D using victual work0

Answered: Image /qna-images/answer/7075e60f-b707-4c8e-91f3-e488186b17d9.jpg

* Ignore Internal Strain Energy due to axial load and shear. A (pinned) 15 ff 2 k/ff V C (hinge) V VO 15 ft 1 1(A) = 1 20 ft -SA EI = constant E = 10,000 ksi I = 2,500 in M.M dx EI EI=Constant E = 10,000 ks: I=2,500 in 4 € (pined) On 6/9 Find Horizontal Deflection at joint 0 using virtual work

* Ignore Internal Strain Energy due to axial load and shear. A (pinned) 15 ff 2 k/ff V C (hinge) V VO 15 ft 1 1(A) = 1 20 ft -SA EI = constant E = 10,000 ksi I = 2,500 in M.M dx EI EI=Constant E = 10,000 ks: I=2,500 in 4 € (pined) On 6/9 Find Horizontal Deflection at joint 0 using virtual work

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Properties of materials

A property of a material describes its characteristics and distinguishes it from the others. The material properties are described it in terms of its hardness, color, appearances, and so on. The material properties generally depend on the type of material, or whether it is a metal or an alloy. There are different kinds of material properties such as mechanical properties, thermal properties, physical properties, and chemical properties. Chemical properties describe the material in terms of a chemical change. The mechanical properties come into existence when the material is acted upon by an external load. Thermal properties are associated with the temperature of the material and specific heat.

Question

100%

A Ignore Internal Strain Energy due to
axial load and shear.
V
A (pinned)
15ff
2 k/ff
V V V
C (hinge)
15 ft
NO
€ (pireed)
1(A) = St
20 ft
EI= constant
E = 10,000 ksi
I = 2,500 in
M₂M
- dx
ΕΙ
Find Horizontal Deflection
EI=Constant
E = 10,000 ks:
I=2,500 in 4
On 6/9
at joint D using victual work
0

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