170 PRACTICAL STRUCTURAL DESIGN 



Pressure on inclined bed: 6 = 16 degrees, n = 1200 x 0.16 = 

 192 Ibs. per sq. in. This is below the allowable safe pressure 

 across the grain, which is 350 Ibs. per sq. in. which we will use. 



_ , , 18.000 _., _ 



Required area in bearing = = 51.5 sq. ins. 



OOU 



Actual area = 10.5 x 8 = 84 sq. ins. The area therefore is more 

 than sufficient for its component of pressure. 



The student is to observe the effect the distribution of the 

 pressure on two bearing faces has on the depth of the cut. It is 

 obvious that when the cut is normal to the stress in the brace 

 the whole thrust must be taken on the toe of the post and none is 

 taken by the inclined face. Actually there is a small component 

 normal to the center line of the brace, but it is so small that it can 

 be neglected. It is probably taken care of by friction of the toe 

 on the cut, or by slight tension in the bolts. 



The small diagram in the upper left-hand corner of Fig. 99 is 

 perhaps self-evident, but will be explained. The load travels 

 down the brace until it reaches a point opposite the center of the 

 inclined face, when it divides, part going to the inclined face and 

 part to the toe. This is drawn to scale on the line A B. The line 

 AC is parallel to the line of the toe. The other lines require no 

 explanation, for the values are marked on the diagram and also 

 on the drawing. 



The vertical component of the inclined face is 18,000 Ibs. and 

 the vertical component of the toe is 10,500 Ibs., the distance 



between them being 4.5 in. The center of gravity = T 



lo,UUU+lU,OU(J 



= 0.369 X 4.5 = 1.66 in. from the right component. 



The bending moment of the couple in the chord = 4 x 25,000 

 = 100,000 in. Ibs. This divided by the reaction gives the distance 

 the center line of the support must be shifted to the left to insure 

 equilibrium. The reaction in this case merely happens to equal 

 the tension in the lower chord, so the center line of the reaction 

 must be 4 ins. to the left of the position above found, or, 4 + 1.66 

 = 5.66 ins. to the left of the center of the lower bed on which the 

 brace rests. 



Fig. 99 purposely contains as few lines as possible in order to 

 avoid confusing the student. In both details there are two inclined 

 bolts with cast-iron washers to keep the pieces in position, so all 

 the bearing areas will have a proper contact. No method is known 



