Beam Deflections - Triangular Load first created 07/05/07 - last modified 11/05/07 Page Author: Ty Harness
Figure 1 - Encastre beam with a triangular load
From figure 1 we can see the load q is distributed as a triangle above the beam and using the equation
of a straight line:
$q(x) = 2/L*q_0 *x$ for $0 < x < L/2$ .....(1)
Neglecting the beams self weight and assuming the reactions Ra and b
are equal then using the equations of statics we can say :
$2*R_a = (q_0/2)*L$ .....(2)
$R_a = (q_0*L)/4$ .....(3)
Figure 2 - Slicing the beam and adding a shear force V and Moment to ensure beam is still in equilibrium
Slicing the beam at x we can write down the bending moment equation to the
left of x where the shearing force V will cause no moment:
$M = (q_0*L)/4*x - 1/2*q*x *(x/3) - M_a$ .....(4)
where x/3 is the centroid of the load triangle taken from x. Noting the problem is statically indeterminate we can't
find Ma from the equations of statics alone and we'll need to use boundary conditions.
Substituting q from equation 1 into equation 5 and simplifying:
and that would mean A = 0. As mentioned, we need another boundary
condition to find Ma and the symmetrical nature of the problem we could assume the slope
at L/2 is also zero which is just still valid for the derived bending moment equation.
