A specimen has circular cross-sectional area with initial diameter d = 26 mm and a gauge length Lo = 200 mm. A force P = 127 kN
1 answer:
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Answer:
a. ε₁=-0.000317
ε₂=0.000017
θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain =3.335 *10^-4
Associated average normal strain ε(avg) =150 *10^-6
θ = 31.71 or -58.29
Explanation:
ε₁=-0.000317
ε₂=0.000017
To determine the orientation of ε₁ and ε₂
θ= -13.28° and 76.72°
To determine the direction of ε₁ and ε₂
=-0.000284 -0.0000335 = -0.000317 =ε₁
Therefore θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain
=3.335 *10^-4
ε(avg) =150 *10^-6
orientation of γmax
θ = 31.71 or -58.29
To determine the direction of γmax
= 1.67 *10^-4
Answer:
hello your question is incomplete attached below is the complete question
answer: There is a hydraulic jump
Explanation:
First we have to calculate the depth of flow downstream of the gate
y1 = ----------- ( 1 )
Cc ( concentration coefficient ) = 0.61 ( assumed )
Yg ( depth of gate opening ) = 0.5
hence equation 1 becomes
y1 = 0.61 * 0.5 = 0.305 m
calculate the flow per unit width q
q = Q / b ----------- ( 2 )
Q = 10 m^3 /s
b = 2 m
hence equation 2 becomes
q = 10 / 2 = 5 m^2/s
next calculate the depth before hydraulic jump y2 by using the hydraulic equation
answer : where y1 < y2 hence a hydraulic jump occurs in the lined channel
attached below is the remaining part of the solution
