What is the factor of safety for a beam designed to support a maximum load of 10
Practice Questions
Q1
What is the factor of safety for a beam designed to support a maximum load of 10 kN if the yield strength of the material is 250 MPa and the beam's cross-sectional area is 50 cm²?
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Questions & Step-by-Step Solutions
What is the factor of safety for a beam designed to support a maximum load of 10 kN if the yield strength of the material is 250 MPa and the beam's cross-sectional area is 50 cm²?
Step 1: Identify the maximum load the beam needs to support, which is 10 kN.
Step 2: Convert the maximum load from kilonewtons (kN) to newtons (N) because 1 kN = 1000 N. So, 10 kN = 10,000 N.
Step 3: Identify the yield strength of the material, which is given as 250 MPa.
Step 4: Convert the yield strength from megapascals (MPa) to pascals (Pa) because 1 MPa = 1,000,000 Pa. So, 250 MPa = 250,000,000 Pa.
Step 5: Identify the cross-sectional area of the beam, which is given as 50 cm².
Step 6: Convert the cross-sectional area from square centimeters (cm²) to square meters (m²) because 1 cm² = 0.0001 m². So, 50 cm² = 0.005 m².
Step 7: Calculate the stress on the beam using the formula: Stress = Max Load / Area. Substitute the values: Stress = 10,000 N / 0.005 m² = 2,000,000 Pa.
Step 8: Calculate the factor of safety using the formula: Factor of Safety = Yield Strength / Stress. Substitute the values: Factor of Safety = 250,000,000 Pa / 2,000,000 Pa.
Step 9: Perform the division: Factor of Safety = 125.
