PINS AND PIN PACKING. 



Bending Moment. The stresses in the members are shown in (c) Fig. 10, which gives the 

 force polygon for the forces. The make-up of the members is shown in (a), and the pin packing 

 on one side is shown in (b). The stresses shown in (c) are applied one-half on each side of the 

 iiicinlMT. the pin acting like a simple beam. The stresses are assumed as applied at the centers 

 of the plates which make the members. 



!>< '$ J r **fey5 



j Him,tPl.< r^t I'*** * ' 



^1 



_ -^ ^^ ^r 



(d) 

 >&*> 



I /^c"' 



j 1-165400*0.55- 07660 



i Vertical Components, 

 \ Moments at 

 1 5=0*" 



<w**m> | - i'^;^r" 



force Di'aqram-^tre55e5 U . ' 5-0*" 



, v Total Moment at 4Z-8 7=I26300<I.S}=23IIOO*" 



- ^08600 * +285000 2 8-176500*3.06 



-15 1600*" -dJ/MxIll-ZdWOO*'' 



3. CALCULATION OF STRESSES IN A PIN. 



FIG. 10. 



Calculation of Stresses in a Pin. The amounts of the forces and the distances between their 

 points of application as calculated from (6) are shown in (d) Fig. 10. The horizontal and vertical 

 components of the forces are considered separately, the maximum horizontal bending moment 

 and the maximum vertical bending moment are calculated forthe same point, and the resultant 

 moment is then found by means of the force triangle. 



In (d) the horizontal bending moments are calculated about the points I, 2, 3, 4; the maximum 

 horizontal moment is to the right of 3, and is 208,600 in.-lb. The vertical bending moments are 

 calculated about points 5, 6, 7, 8; the maximum bending moment is to the right of 8, and is 

 283,000 in.-lb. The maximum bending moment is at, and to the right of 4 and 8, and is, M = 



+ 283,000* = 351,600 in.-lb. Substituting in the formula,/ = M-c/I, the maximum 

 bending stress is / = 16,600 Ib. per sq. in. The allowable bending stress in pins for which this 

 bridge was designed was i8,oop Ib. per square inch. The allowable bending moments on pin 

 are given in Table 98. 



Shear. The shear is found for both the horizontal and vertical components as in a simple 

 beam, and is equal to the summation of all the forces to the left of the section. The maximum 

 horizontal shear is between I and 2, and is 165,400 Ib. The shear between 2 and 3 is 165,400 

 99,300 = 66,100 Ib. The maximum vertical shear is between 6 and 7, and is 126,300 Ib. The 

 resultant shear between 2 and 3, and 6 and 7, is, V = ^i 26,300* + 66,100* = 145,000 Ib., which 

 is less than the horizontal shear between I and 2. The maximum shear, therefore, comes 





