COMPOUND STRAINS AND STRESSES 149 
is W(z+7?), and if /, is the maximum stress, then Wx+ Wi = }bd’/,. 
Hence Wt = }ld*( 7, -/, 
The distribution ri ‘the stresses due to bending on the upper half of 
the sections YY and Y’Y’ is shown in Fig. 200. A portion A of the 
beam bounded by the sections YY, Y’Y’, the top surface of the beam, and 
a horizontal section at a distance h above the neutral surface, is pulled to 
2 
the right by a force R, = wA($ : =), and it is also pulled to the left by a 
- force R,= TAC -*). The resultant of these two pulls is a pull to the 
_ left by a force R=R, - R, =0(f,—J;) (F= =) and this is balanced by 
the horizontal shear on the under surface of A. Let ¢ equal the intensity 
2 
_ of the shear stress on the under surface of A, then btg = b( /, - -A)(- a 
But £,-A= ; Sy therefore 7 = ul ae 3 
be the intensity of the transverse shear stress on the section YY at a 
distance h from the neutral axis. 
6W (¢- h? 
bP\4 a 
_ parabola which, when drawn as shown in Fig. 201, represents the dis- 
_ tribution of the shear stress on a transverse section of the beam. 
The maximum shear stress in the case just considered evidently 
occurs at the neutral surface of the beam or neutral axis of the transverse 
section where its intensity is pus , but since the total 
), and this by Art. 141 must 
The equation g = =) connecting g and h is the equation to a 
; — shear is W,theaverage transverse shear stress 
8 70! , hence the maximum transverse shear stress is 14 
times the mean. 
Proceeding now to the general case of a beam 
_ of any form of cross section, and referring to 
Figs. 200 and 201, and also to Fig. 202, which 
comet the section of the beam, the stress due Fig. 202. 
to bending at a distance y from the neutral axis 
=f “n at the section YY, and f = 1 at the section Y’Y’. 
% 
p= Lfedy = Sf, : “oy = “1 Lyzby between the limits y=h and y=y.. 
‘But ae ay), where @ is the area of the section beyond the line at a dis- - 
y tance h from the neutral axis, and y, is the distance of the centre of gravity 
of that area from the neutral axis. Therefore R, =fray,. In like manner 
i . 
Pn, ~7 to: Hence R=R,— R, =(f,- sy Again, Wz “a avail 
“ : 
4 W(x +?) =f, therefore Wt = = fy-f,). Also, R= qbt, Hence gq = Le 3 
x | ; iN 
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