Question

Hello, please help me with this homework problem. Please show all work. (Correct answers are listed at the bottom)

Homework Assignment #33 10 points The University of Akron Department of Civil Engineering 4300 202: Introduction to Mechanics of Solids 1. A solid tube is subjected to the loading shown. The tube has a diameter of 1.5 in. 8 in. (a) Determine which internal force quantities would cause normal stresses and which would cause shear stresses on a cross-section cut through points A and B 600 lb (b) Determine the state of stress at point A (c) Determine the state of stress at point B in. 500 lb Answers: (a) Normal: Ny, Mx, M., Shear: Ty, Vx, V 800 lb (b) ?-0, ?,-4 1.5 ksi (tension), ?--0, txy-0 Tx,-0, ???= +14.5 ksi (combination of Ty and V,) (c) ?.-0, ?y-14.0 ksi (compression), ?,-0, txy--I 3.9 ksi (combination of T, and

Answer 1

a)

the cut section is shown below. Stresses are identified.

b) State of stress at point A and B.

From sum of forces in x-direction = F_x =0

V_x - 500 lb = 0

V_x = 500 lb.

The moment about x is zero.

M_x + 600 lb*8 =0.

M_x = -4800 lb-in

Force in y-direction is zero.

Ny + 800 lb = 0.

N_y = -800 lb.

The moment about y is zero.

M_y - 600*12 = 0

M_y = 7200 lb-in

Sum of forces in Z-direction is zero.

V_z + 600 lb =0.

V_z = -600 lb.

The moment in Z-direction is zero.

M_z = 0

M_z + 800 * 12 +500*8 = 0

M_z = -9600 -4000 = -13600 lb-in

Stress at point B = $\tau _{B}=vQ/Id$

I = moment of inertia = 3.14*d⁴ / 64 = 3.14*1.5⁴ / 64 = 0.2483 in³

d = 1.5 in

Q=Ay

A = semi circle area = (3.14/2) * (0.75)² = 0.883125 in²

y = centroid of semi circle = (4/3*pi) * radius = (4/3*3.14) * 0.75 = 0.3184 in

Shear stress = $\tau _{B}=vQ/Id$ = (500*0.883125*0.3184) / (0.2483*1.5) = 377 psi

Shear stress = $\tau _{yz}$ = Mz / I = 4800*0.75 / 0.2483 = 14.5 ksi

Normal stress = $\sigma _{b}=N/A$ = 800 / pi*r² = 800 / 3.14*0.75² = 0.452 ksi

$\tau _{B}$ = 7200*0.75 / J = 7200*0.75 / (pi/2) * r⁴ = 7200*0.75 / ((3.14/2)*0.75⁴) = 10870 psi