If a factor of safety =1.9, what would be the allowable stress?
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If a factor of safety =1.9, what would be the allowable stress?
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- Compute the value of stress Tensile load=22 N Cross sectional area=2 mm2A steel rod with a cross-sectional area of 188 mm2 is stretched between two fixed points. The tensile load at 23°C is 6,332 N. What will be the stress at -15°C ? Assume a = 11.7 x 10-6 m/(m.°C) and E = 198 GPa. Answer must be in MPa.Define the term neutral stress.
- Problem: A steel rod with a cross-sectional area of 180 mm2² is stretched between two fixed points. The tensile load at 25°C is 3,600N. Assume thermal coefficient equal to 11.7 x 106 mm/ (mm°C) and E= 200 GPa. a. What is the temperature at which the stress will be zero? b. What is tthe temperature at which the compressive stress will be 48 MPa?(7).!! Show Complete & Detailed Solution w/ FBD !!The bronze bar 3 m long with a cross-sectional area of 350 mm^2 is placed between two rigid walls. At a temperature of –25C, there is a gap = 2.5 mm, as shown in the figure. Find the temperature at which the compressive stress in the bar will be 30 MPa. Use α = 18.0 x 10–6 /C and E = 80 GPa.
- If you add a diagram it would be well and good.A mild steel bar of square cross-section 40 mm x 40 mm is 400 mm long it is subjected to a longitudinal tensile stress of 440 N/mm² and lateral compressive stress is 200 N/mm² in perpendicular directions. E = 2 x 105 N/mm², μ = 0.3. What is the approximate elongation of the bar in the longitudinal direction?Q no.1