Answered: by the analytical method and Mohr’s…

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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by the analytical method and Mohr’s circle
Find ε_x' , ε_y' and γ_x'y'
Find main directions θp and main values ε₁ and ε₂

200MS
30 MS
GOOMS
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Expert Solution

Step 1 Given

$ε_{x} = - 600 μ s ε_{y} = 200 μ s γ_{x y} = 30 μ s$

Step 2 Calculation of strains when plane is rotated

$ε_{x}' = \frac{ε_{x} + ε_{y}}{2} + \frac{ε_{x} - ε_{y}}{2} \cos 2 θ + \frac{γ_{x y}}{2} \sin 2 θ ε_{x}' = \frac{- 600 + 200}{2} + \frac{- 600 - 200}{2} \cos (2 \times 30) + \frac{γ_{x y}}{2} \sin (2 \times 30) ε_{x}' = - 200 - 400 \cos 60 + \frac{30}{2} \sin 60 ε_{x}' = - 200 - 200 + 15 \times 0.866 ε_{x}' = - 387.01 ε_{y}' = \frac{ε_{x} + ε_{y}}{2} - \frac{ε_{x} - ε_{y}}{2} \cos 2 θ - \frac{γ_{x y}}{2} \sin 2 θ ε_{y}' = \frac{- 600 + 200}{2} - \frac{- 600 - 200}{2} \cos (2 \times 30) - \frac{30}{2} \sin (2 \times 30) ε_{y}' = - 200 + 400 \cos 60 - \frac{30}{2} \sin 60 ε_{y}' = - 200 + 200 - 15 \times 0.866 ε_{y}' = - 13 \frac{γ_{x' y'}}{2} = - \frac{ε_{x} + ε_{y}}{2} \sin 2 θ + \frac{γ_{x y}}{2} \cos 2 θ \frac{γ_{x' y'}}{2} = - \frac{- 600 + 200}{2} \sin 60 + \frac{30}{2} \cos 60 \frac{γ_{x' y'}}{2} = 200 \sin 60 + 15 \cos 60 \frac{γ_{x' y'}}{2} = 173.21 + 7.5 γ_{x' y'} = 361.42$

Step 3 Calculation of principal strains

$σ_{1, 2} = \frac{ε_{x} + ε_{y}}{2} \pm \sqrt{{(\frac{ε_{x} - ε_{y}}{2})}^{2} + {(\frac{γ_{x y}}{2})}^{2}} σ_{1, 2} = \frac{- 600 + 200}{2} \pm \sqrt{{(\frac{- 600 - 200}{2})}^{2} + {(\frac{30}{2})}^{2}} ε_{1, 2} = - 200 \pm \sqrt{{(400)}^{2} + {(15)}^{2}} ε_{1, 2} = - 200 \pm 400.28 ε_{1} = - 600.28 ε_{_{2}} = 200.28$

$\tan 2 θ_{p} = \frac{γ_{x y}}{ε_{x} - ε_{y}} \tan 2 θ_{p} = \frac{30}{- 600 - 200} 2 θ_{p} = - 2.148 θ_{p} = - 1.074$

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