The axle of an automobile is acted upon by the forces and couple shown. knowing that the diameter of the solid axle is 32 mm, determine (a) the principal planes and principal stresses at point h located on top of the axle, (b) the maximum shearing stress at the same point. use mohr's circle

First we need to convert the mm to inches to make our computation easier.

1mm = 0.0393701

32mm * 0.0393701 = 1.25 in

Solution:

C = 1/2d = ½ (1.25) = 0.625 in^4

Tension: tension = Te/J = 2T/ piC^3

= (2)(2500)/pi (0.0625)^3 = 6.519 x 10^3 psi = 6.519 ksi

Bending:

I = pi/4 * c^4 = 119.842 x 10^-3 in^4

M = (5)(600) = 3600 lb in

G = My/I = (3600)(0.625)/119.842 x 10^-3 = -18.775 x 10^2 psi = -18.775ksi

Gx = -18.775 ksi

Gy = 0

Txy = 6.519 ksi

G ave – ½ (Gx + Gy) = -9.387 ksi
R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2 + (6.519)^2 = 11.429 ksi

1.       G1 = Gave + R = -9/387 + 11.429 = 2.04 ksi

G2 = Gave - R = -9/387 - 11.429 = -20.8

Tan 2ϴp = 2txy/Gx – Gy = 2(6.519)/-9.387 = -1.3889

ϴp = -27.1 degrees and 62.9 degrees

2.       Tmax = R = 11.43 ksi

R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2 + (6.519)^2 = 11.429 ksi

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)