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Safety Factor for Simple Loading

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Prequisite Knowledge
Moment • Safety Factor

Problem
Link AC is 1/8" thick, has width of 1/2", and is made with a steel that has an ultimate strength of 70 ksi in tension. What is the safety factor for member AC if the structure shown below was desiged to support a 2-kip load P?


Solution

SF =	ultimate load allowable load

You are given the force P acting downward. This is 2 kips. So you want to find the tensile force that load is causing in member AC.

This is found using a moment equation for member BD.

It helps to draw a FBD of the member. Doing so will present the force in need.


Angle θ is found by the relation:

cos θ = a/h, where

θ = cos^-1(12/15) = 36.87°

Then length BD is found by using the relation:

cos θ = a/h, where h = length BD and a = length 15 in,

L_BD = 15 / (cos 36.87°) = 18.75 in

An equation can then be written for the moment about B.

ΣM_B = 0 = F_AC(12) - (2*cos36.87°)(18.75)

F_AC = 2.5 lb = the allowable load

The ultimate load = (the ultimate stress)*(the cross sectional area), so

F_u = (70 ksi)(0.125*.5) = 4.375 kip

Therefore, the safety factor used for the member is,

SF =

4.375 kip 2.5 kip

= 1.75

SF = 1.75