ABSTRACT
FIGURE 6.14: Shearing forces and moments on a cantilever beam with a tip mass at the free end.
as depicted in Figure 6.15. A general force balance yields∫ x+∆x x
f(t, s) ds+ S(t, x+ ∆x) − S(t, x)− fI
− ∫ x+∆x x
γd ∂y
∂t (t, s) ds = 0, (6.2)
where fI represents the inertia force due to the mass of the differential element abdc and is given by a sum of point inertia forces fi
fI = ∫ x+∆x x
fi(t, s) ds = ∫ x+∆x x
ρ ∂2y
∂t2 (t, s) ds.
