Answer:
1st part: Section W18X76 is adequate
2nd part: Section W21X62 is adequate
Explanation:
See the attached file for the calculation
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
Express it in standard form and apply the basic indices laws to simplify
Answer:
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Explanation:

Answer:
Pressure = 11.38 psi
Force = 13.981 Ibf
Explanation:
Step by step solution is in the attached document.