Question 10.1: Figure 10–12 shows an L-shaped lever carrying a downward for...

Objective   Compute the stress condition and draw the stress element.

Given          Geometry and loading in Figure 10–12; P = 1500 N

Dimensions: a = 150 mm, b = 300 mm, D = 30 mm

Analysis      Figure 10–13 shows the lever separated into two free-body diagrams. The point of interest is labeled “element K” at the right end of the lever where it joins the fixed support. The element is subjected to bending stress due to the reaction moment at the support, M_{2} . It is also subjected to torsional shear stress due to the torque, T.

Results        Using the free-body diagrams in Figure 10-13,  we can show that

M_{1} = T = Pa = (1500 N)(150 mm) = 225 000 N·mm

M_{2} = Pb = (1500 N)(300 mm) = 450 000 N·mm

Then the bending stress on top of the bar, shown as element K in Figure 10–13(b), is

\sigma_{x} = \frac{M_{2}c}{I} = \frac{M_{2}}{S}

The section modulus S is

S = \frac{ \pi D^{3}}{32} = \frac{ \pi(30  mm)^{3}} {32} = 2651 mm³

Then

\sigma_{x} =  \frac{ 450  000  N·mm}{2651  mm^{3}} =  169.8 N/mm² = 169.8 MPa

The torsional shear stress is at its maximum value all around the outer surface of the bar, with the value of

\tau = \frac{T}{Z_{p}}

The polar section modulus Z_{p} is

Z_{p} = \frac{ \pi D^{3}}{16} = \frac{ \pi(30  mm)^{3}} {16} = 53011 mm³

Then

\tau =  \frac{ 225  000  N·mm}{5301  mm^{3}} =  42.4 N/mm² = 42.4 MPa

Comment  The bending and shear stresses are shown on the stress element K in Figure 10–14. This is likely to be the point of maximum combined stress, which will be discussed later in this chapter. At a point on the side of the bar on the y-axis, a larger shearing stress would exist because the maximum torsional shear stress combines with the maximum vertical shear stress due to bending. But the bending stress there is zero. That element should also be analyzed.