Question 10.3: A picnic table in a park is made by supporting a circular to...

Objective  Compute the maximum stress in the tube in Figure 10–19 and the design factor based on both yield strength and ultimate strength.

Given         Loading and tube dimensions shown in Figure 10–19

Load is the force due to a 135 kg mass on the edge of the table.

Tube is aluminum, 6061-T4; s_{y} = 145 MPa; s_{u} = 241 MPa (Appendix A–14).

Section properties for the tube from Appendix A–9(d), ref r.

S = 5.909 \times 10^{4} mm³  A = 1963 mm²

Analysis    The tube is subjected to combined bending and direct compression as illustrated in Figure 10–20, the free-body diagram of the tube. The effect of the load is to produce a downward force on the top of the pipe while exerting a moment in the CW direction.

The moment is the product of the load times the radius of the table top. The reaction at the bottom of the tube, supplied by the concrete, is an upward force combined with a CCW moment. Use the Guidelines for Solving Problems with Combined Normal Stresses.

Results       Step 1. Figure 10–20 shows the free-body diagram. The force is the gravitational attraction of the 135 kg mass:

F = m · g = (135 kg)(9.81 m/s²) = 1324 N

Step 2. No forces act at an angle to the axis of the tube.

Step 3. Now the direct axial compressive stress in the tube is

\sigma_{a} = – \frac{F}{A} = – \frac{1324  N}{1963  mm²} = -0.674 MPa

This stress is a uniform, compressive stress across any cross section of the tube.

Step 4. No forces act perpendicular to the axis of the tube.

Step 5. Since the force acts at a distance of 1.1 m from the axis of the tube, the moment is

M  = F · R  =(1324 N)(1.1 m) = 1456  N·m .

The bending stress computation requires the application of the flexure formula:

\sigma_{b} = \frac{Mc}{I} = \frac{M}{S}

where S = 5.909 × 10^{4}  mm^{3}

Then,

\sigma_{b} = \frac{1456 N·m}{5.909 × 10^{4}  mm^{3}} \times \frac{10^{3}  mm}{m} = 24.64 MPa

Step 6. The bending stress \sigma_{b} produce compressive stress on the right side of the tube and tension on the left side. Since the direct compression stress adds to the negative bending stress on the right side, that is where the maximum combined stress would occur. Using Equation 10–1, the combined stress would then be

\sigma_{c} =-\sigma_{a}  -\sigma_{b}  = (-0.674 – 24.64) MPa = -25.31 MPa

Step 7. The design factor based on yield strength is

N = \frac{s_{y}}{\sigma_{c}} = \frac{145  MPa}{25.31  MPa} = 5.73

The design factor based on ultimate strength is

N = \frac{s_{u}}{\sigma_{c}} = \frac{241  MPa}{25.31  MPa} = 9.52

Comments  These values for N should be compared with the design stress values listed in Table 3–2. There it is suggested that N = 2 based on yield strength for static loading and N = 12 based on ultimate strength for impact. If the person simply sits on the edge of the table, the loading would be considered to be static and the stress is acceptable. But a person jumping on the edge would apply an impact load, and that may happen in a public park environment. The design factor of 9.52 is unacceptably low. Therefore, a larger tube should be used.