-->//(Design against Static Load) Example 4.18
 
-->//Refer Fig.4.57 on page 126
 
-->//Tensile yield strength of FeE250 Syt (N/mm2)
 
-->Syt = 250
 Syt  =
 
    250.  
 
-->//Factor of safety fs
 
-->fs = 5
 fs  =
 
    5.  
 
-->//Diameter of bars to be sheared D (mm)
 
-->D = 6.25
 D  =
 
    6.25  
 
-->//Ultimate shear strength of the material Sus (N/mm2)
 
-->Sus = 350
 Sus  =
 
    350.  
 
-->//Permissible bearing pressure on the pins p (N/mm2)
 
-->p = 10
 p  =
 
    10.  
 
-->//Pin length to diameter ratio r1
 
-->r1 = 1.25
 r1  =
 
    1.25  
 
-->//Link cross-section width to thickness ratio r2 
 
-->r2 = 2
 r2  =
 
    2.  
 
-->//Distance between pinA and pinB l1 (mm)
 
-->l4 = 400
 l4  =
 
    400.  
 
-->//Distance between bar to be sheared and pinA & force application point and pinC l2 (mm)
 
-->l2 = 100
 l2  =
 
    100.  
 
-->//Distance between the force applied and pinD l3 (mm)
 
-->l3 = 1000
 l3  =
 
    1000.  
 
-->//Thickness of gunmetal bush over pin C t (mm)
 
-->t = 2.5
 t  =
 
    2.5  
 

Diameter of pins(d1) = 15.000000 mm

Length of pins(l1) = 20.000000 mm

Diameter of link(d) = 10.000000 mm

Width of lever cross-section(h) = 40.000000 mm

Thickness of lever cross-section(b) = 20.000000 mm

The design is safe