• So, with the following effective stress area estimates.. As = 0.1414 in2 Ats = 0.3458 in^2 …the predicted tensile load that results in a safety factor of 1.25 is… • For thread shear: Fts = Ats*1/sf* y/2 = 0.3458*.8*23206/2 = 3209.9 lbf • For tensile failure: Ft = As*1/sf* y = 0.1414*.8*23206 = 2625.1 lbfGet A Quote
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The bending stress is zero at the neutral axis and varies linearly across the cross section from maximum compressive to maximum tensile. SL = Mb * c/I Where, Mb = Moment of the beam. c = distance of point of interest from the neutral axis.Get A Quote
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