176 PRACTICAL STRUCTURAL DESIGN 



Depth of toe, = 45. n = 540 Ibs. per sq. in. 

 Required area for end bearing = --r - = 46.4 sq. ins. 



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464 

 Required depth of vertical cut = ~ = 5.8 ins. 



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Required area of horizontal cut, = 45. n = 540 Ibs. per sq. in. 



Use vertical component of the diagonal load, which for 45 de- 



grees is one-half. 



nn n . 

 = 33.3 sq. ins. 



n 



2 x 540 



33.3 



Length of horizontal cut = 5 ^r = 5.5 ins. 



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In this example the angle is 45 degrees and the vertical cut 

 and horizontal cut will be equal, each being 8 x 0.707 = 5.66 ins. 

 The vertical cut, for the assumed fiber stress, should be 5.8 ins., 

 which indicates that the casting should be designed with the 

 two bearing surfaces forming an angle differing enough from 

 90 degrees to keep the stress within the limits fixed. If the design 



is not altered the stress on the end area = -=-^ - - = 552 Ibs. per 



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sq. in., which is an increase of only 4 per cent, so will be allowed 

 to stand. 



The maximum unit bearing pressure of the shoe on the lower 

 chord must not exceed the allowable bearing stress on the side 

 of the fibers. Draw a horizontal line from the mid-height of the 

 toe to intersect with the diagonal line meeting the point of the 

 toe. From this intersection drop a vertical line to represent 

 the vertical component of the thrust. The diagonal line rep- 

 resents the diagonal thrust and the horizontal and vertical 

 components, respectively, are shown as heavy arrows. 



The distance from the vertical component to the front edge 

 of shoe (point of maximum pressure) scales 8.5 ins. and this length 

 will be called a. Next find the length of the shoe. The lugs each 

 carry one-half the shear, or 12,500 Ibs. 



12 500 

 The area for bearing = ' nn = 10.4 sq. ins. The depth of the 



lug = - = 1.3 ins. (make it 1.5 ins.). 



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Length required for shear = 5 -T = 13 ins. 



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