Question 7.11: Figure 7–32 shows the cross section of a beam that is to be ...
Figure 7–32 shows the cross section of a beam that is to be made from malleable iron, ASTM A220, grade 80002. The beam is subjected to a maximum bending moment of 1025 N · m, acting in a manner to place the bottom of the beam in tension and the top in compression. Compute the resulting design factor for the beam based on the ultimate strength of the iron. The area moment of inertia for the cross section is 1.80 × 10^{5} mm^{4} .
Objective Compute the design factor based on the ultimate strength.
Given Beam shape shown in Figure 7–32; I = 1.80 × 10^{5} mm^{4} ; M = 1025 N · m Material is malleable iron, ASTM A220, grade 80002.
Analysis Because the beam cross section is not symmetrical, the value of the maximum tensile stress at the bottom of the beam, σ_{tb} , will be lower than the maximum compressive stress at the top, σ_{ct} . We will compute
σ_{tb} = Mc_{b} /I and σ_{ct} = Mc_{t} /I
where c_{b} = \bar{Y} = 14.04 mm and c_{t} = 50 − 14.04 = 35.96 mm. The tensile stress at the bottom will be compared with the ultimate tensile strength to determine the design factor based on tension, N_{c} , from
σ_{tb} = s_{u} /N_{t} and N_{t} = s_{u} / \sigma_{tb}
Where s_{u} = 655 MPa from Appendix A–13. Then the compressive stress at the top will be compared with the ultimate compressive strength to determine the design factor based on compression, N_{c} , from
| A–13 Typical properties of cast iron.a | |||||||||||
| Ultimate strength | Yield strength | ||||||||||
| s_{u}^{b} | s_{uc}^{c} | s_{us}^{c} | s_{ut}^{c} | Modulus of elasticity. E^{c} | |||||||
| Meterial type and grade | ksi | MPa | ksi | MPa | ksi | MPa | ksi | MPa | psi | GPa | Percent elogation |
| Gray iron ASTM A48 | |||||||||||
| Grade 20 | 20 | 138 | 80 | 552 | 32 | 221 | – | – | 12.2 \times 10^{6} | 84 | <1 |
| Grade 40 | 40 | 276 | 140 | 965 | 57 | 393 | – | – | 19.4 \times 10^{6} | 134 | <0.8 |
| Grade 60 | 55 | 379 | 170 | 1170 | 72 | 496 | – | – | 21.5 \times 10^{6} | 148 | <0.5 |
| Ductile iron ASTM A536 | |||||||||||
| 60-40-18 | 60 | 414 | – | – | 57 | 393 | 40 | 276 | 24 \times 10^{6} | 165 | 18 |
| 80-55-6 | 80 | 552 | – | – | 73 | 503 | 55 | 379 | 24 \times 10^{6} | 165 | 6 |
| 100-70-3 | 100 | 690 | – | – | – | – | 70 | 483 | 24 \times 10^{6} | 165 | 3 |
| 120-90-2 | 120 | 727 | 180 | 1240 | – | – | 90 | 621 | 23 \times 10^{6} | 159 | 2 |
| Austempered ductile iron (AID) | |||||||||||
| Grade 1 | 125 | 862 | – | – | – | – | 80 | 552 | 24 \times 10^{6} | 165 | 10 |
| Grade 2 | 150 | 1034 | – | – | – | – | 100 | 690 | 24 \times 10^{6} | 165 | 7 |
| Grade 3 | 175 | 1207 | – | – | – | – | 125 | 862 | 24 \times 10^{6} | 165 | 4 |
| Grade 4 | 200 | 1379 | – | – | – | – | 155 | 1069 | 24 \times 10^{6} | 165 | 1 |
| Malleable iron ASTM A220 | |||||||||||
| 45008 | 65 | 448 | 240 | 1650 | 49 | 338 | 45 | 310 | 26 \times 10^{6} | 170 | 8 |
| 60004 | 80 | 552 | 240 | 1650 | 65 | 448 | 60 | 414 | 27 \times 10^{6} | 186 | 4 |
| 80002 | 95 | 655 | 240 | 1650 | 75 | 517 | 80 | 552 | 27 \times 10^{6} | 186 | 2 |
σ_{ct} = s_{us} /N_{c} and N_{c} = s_{uc} / \sigma_{ct}
where s_{uc} = 1650 MPa from Appendix A–13. The lower of the two values of N will be the final design factor for the beam.
Results At the bottom of the beam,
σ_{tb} = \frac{Mc_{b}}{I} = \frac{(1025 N·m)(14.04 mm)}{1.80 \times 10^{5} mm^{4}} \frac{(1000 mm)}{m} = 79.95 MPa
N_{c} = s_{us} / \sigma_{tb} = 655 MPa/79.95 MPa = 8.19
At the top of the beam,
σ_{tb} = \frac{Mc_{t}}{I} = \frac{(1025 N·m)(35.96 mm)}{1.80 \times 10^{5} mm^{4}} \frac{(1000 mm)}{m} = 204.8 MPa
N_{c} = s_{us} / \sigma_{ct} = 1650 MPa/204.8 MPa = 8.06
Comment The compressive stress at the top of the beam is the limiting value in this problem because the smaller design factor exists there. Note, however, that the two values of the design factor were quite close to being equal, indicating that the shape of the cross section has been reasonably well optimized for the different strengths in tension and compression.