Give the properties table below. A992 Steel 7178 T76 Aluminum Yield Strength 50 ksi 73 ksi Ultimate strength 75 ksi 85 ksi Modulus of Elasticity 29,000 ksi 10,500 ksi Assume you are designing a rectangular member with a specific allowed height. For a design where deflection criterion controls which material would require a larger cross section? 6. Give the For a design where tension stress controls, which material requires the larger cross section?
The Correct Answer and Explanation is :
When designing a rectangular member with a specific allowed height, the choice between A992 steel and 7178-T76 aluminum depends on whether the design is governed by deflection criteria or tensile stress criteria.
Deflection Criterion:
Deflection in a beam is influenced by its material’s modulus of elasticity (E). A higher modulus indicates a stiffer material that will deflect less under a given load. The flexural rigidity of a beam is given by the product EI, where I is the moment of inertia of the cross-sectional area. For a rectangular cross-section with a fixed height (h), I is proportional to the width (b) and the cube of the height (I = b*h³/12). Since h is constant, I is directly proportional to b. Therefore, to achieve the same deflection, the required width b is inversely proportional to E.
Given:
- A992 Steel: E = 29,000 ksi
- 7178-T76 Aluminum: E = 10,500 ksi
The ratio of the moduli is:
[
\frac{E_{\text{Steel}}}{E_{\text{Aluminum}}} = \frac{29,000}{10,500} \approx 2.76
]
This means steel is approximately 2.76 times stiffer than aluminum. To achieve the same deflection, the aluminum member would need to have a cross-sectional area about 2.76 times larger than that of the steel member. Therefore, under deflection-controlled design, 7178-T76 aluminum would require a larger cross-section compared to A992 steel.
Tension Stress Criterion:
When tensile stress is the controlling factor, the design must ensure that the material’s yield strength is not exceeded. The required cross-sectional area (A) is determined by the applied force (F) divided by the material’s yield strength (σ_y):
[
A = \frac{F}{\sigma_y}
]
Given:
- A992 Steel: σ_y = 50 ksi
- 7178-T76 Aluminum: σ_y = 73 ksi
The ratio of the yield strengths is:
[
\frac{\sigma_{y,\text{Aluminum}}}{\sigma_{y,\text{Steel}}} = \frac{73}{50} = 1.46
]
This indicates that 7178-T76 aluminum has a 46% higher yield strength than A992 steel. Consequently, for the same applied tensile force, the aluminum member can have a smaller cross-sectional area compared to the steel member. Therefore, under tension stress-controlled design, A992 steel would require a larger cross-section than 7178-T76 aluminum.
Conclusion:
In summary, if deflection criteria control the design, 7178-T76 aluminum would necessitate a larger cross-sectional area due to its lower modulus of elasticity. Conversely, if tensile stress criteria control the design, A992 steel would require a larger cross-sectional area because of its lower yield strength compared to 7178-T76 aluminum.