117 
BEAM OF RECTANGULAR CROSS-SECTION UNDER vlNY SYSTEM OF LOAD. 
Sdiiib rougli Gxp8iiiiiBiits on a block of india-rubbGr lying on a woocIgii tabl© liavG 
confiiniBcl the lesult that the block is lifted, out of contact with the table away from 
the load, and that the area of contact is of the above order. 
PART HI. 
Solution for a Beam Under Asymmetrical Normal Forces; Special Case 
OF Two Opposite Concentrated Loads not in the same Vertical 
Straight Line. 
§ 2/. Expressions for the Displacements and Stresses in Series. 
Let us now proceed to consider what the general solution becomes in the case of a 
beam subject to normal forces which are now no longer restricted to lie symmetrical. 
In this case coefficients y and S come in, as well as a, / 3 ; k, p, 0 being all zero. 
Considei particularly a lieam (fig. v.) sulject to a downwards concentrated 
load W, acting upon its upper surface at a; = /, and an upwards concentrated 
load W, acting upon its lower surface at a; = — 1 . 
Such a system by itself is not in equilibrium. But the solution will introduce 
two shears over the ends, equal to 2 (y„ - Z,) by equation ( 50 ). 
In the case taken above a, = ^ W/ 2 a, == A, = - -cos -S, 
a ' ' " 
_ __ W . 
— sin ml, where m — n^rrla, n being an integer. Hence the shears over the 
I jiTT a ~ U ’ these will satisfy the conditions of rigid 
equilibrium. 
V/e then find the following expressions for the stresses and displacements in series : 
