Normal Stress

 

 

Example:  (continued)                                          

 

 

                         

 

 

For equilibrium of the plate:   CCW  ΣMB = 0    - Ax(12) + 300(6)  =  0      Ax = 150.0 lb

 

 

Note from statics that the link AC is a two-force member.  So that it is either in tension or

in compression.  In this case link AC carries a tensile force, R, directed along the link.

 

For equilibrium of  link AC:    ΣFx  =  0      -  R cos(30)  + 150.0  =  0        R  =  173.2 lb

 

 

Cross-sectional area of link  =  Alink = (1/4)(1/16)  =  1/64 in2

 

 

So the normal stress in link AC  =  σlink = 173.2/(1/64)  =  11085psi =  11.09 ksi   (result)

 

Click here for another example.

 

                                                 

 


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