Question 7.7: A support beam for a conveyor system for a production line c...

Objective   Specify a suitable steel. ‘

Given          Loading pattern in Figure 7–20; loads are repeated.

Beam is to be circular, D = 50 mm.

Analysis     Use the type B procedure given in this section.

Results       Step 1. Figure 7–21 shows the completed shearing force and bending moment diagrams. The maximum bending moment is 2.00 kN · m at the support at C.

Step 2. Appendix A–1 gives the formula for S for a round bar.

S = πD³/32 = π(50 mm)³ /32 = 12 272 mm³

Step 3. Using Equation (7–5),

\sigma_{max} = \frac{M}{S} = \frac{2.0  kN·m}{12  272  mm^{3}} \times \frac{10^{3}  mm}{m} \times \frac{10^{3} N}{kN}

\sigma_{max} = 163 N/mm² = 163 MPa

Step 4. We can use Table 7–1 to determine an appropriate formula for design stress. The steel selected should be highly ductile because of the repeated load. Then, we will use

\sigma_{d} = s_{u} / 8

Step 5. Let \sigma_{max}  = 163 MPa = \sigma_{d} = s_{u} / 8

Step 6. Solving for s_{u} gives

s_{u} = 8(\sigma_{max}) = 8(163 MPa) = 1304 MPa

Step 7. It was decided to use steel.

Step 8. Appendix A–10 lists several common steel alloys. From that table, we can select candidate materials that have good ductility and an ultimate strength of at least 1304 MPa. Four are listed as follows:

SAE 1080 OQT 700; s_{u} = 1303 MPa; 12% elongation

SAE 1141 OQT 700; s_{u} = 1331 MPa; 9% elongation

SAE 4140 OQT 700; s_{u} = 1593 MPa; 12% elongation

SAE 5160 OQT 900; s_{u} = 1351 MPa; 12% elongation

Step 9. For applications to beams carrying repeated loads, it is typical to use a medium carbon steel. Either the SAE 4140 or the SAE 5160 could be used. With 12% elongation, ductility should be adequate.

Comment   Note in Appendix A–10 that SAE 4140 OQT 900 has an ultimate strength of 1289 MPa and 15% elongation. The strength is within 2% of the computed value. It may be suitable to specify this material to gain better ductility. A slight reduction in the design factor would result. Because the values in Table 7–1 are somewhat conservative, this would normally be justified. Alternatively, a larger diameter bar could be used resulting in a lower design stress. It may then be possible to use a lower cost steel.