title: Fracture Mechanics: Fundamentals and Applications, Third Edition
author: Ted L. Anderson
description: An audience of engineers, researchers, and students will appreciate this publication as a reference tool with theoretical background and practical applications.<BR> MCEER, 2005
zlib_text: Chapter 12 
   
   
  Page 1 
  CHAPTER 12 – Static Equilibrium; Elasticity and Fracture 
   
  1. 
  From the force diagram for the sapling we can write  
   
   
  ?Fx = F1 – F2 sin 20° – F3 cos α = 0; 
   
   
   
  380 N – (255 N) sin 20° – F3 cos α = 0,  or   
   
   
   
  F3 cos α = 293 N. 
   
   
  ?Fy = F2 cos 20° – F3 sin α = 0; 
   
   
   
  F3 sin α = (255 N) cos 20° =  240 N. 
   
  Thus we have 
   
   
  F3 = [(293 N)2 + (240 N)2]1/2 =        379 N. 
   
   
  tan α = (240 N)/(293 N) = 0.818,  α = 39.3°.   
   
  So  θ = 180° – α =        141°. 
   
   
   
  2. 
  From the force diagram for the junction we can write  
   
   
  ?Fx = F2 – F1 sin 45° = 0. 
   
  This shows that F1 > F2 , so we take F1 to be the maximum. 
   
   
  ?Fy = F1 cos 45° – Mg = 0; 
   
   
  Mg = (1150 N) sin 45° =       813 N. 
   
   
   
   
   
   
   
  3. 
  We choose the coordinate system shown, with positive torques  
   
  clockwise.  For the torque from the person’s weight about the  
   
  point B we have 
   
   
  τB = MgL = (56 kg)(9.80 m/s2)(3.0 m) =        1.6 × 103 m · N. 
   
   
   
   
  4. 
  We choose the coordina