Jerehmy
Structural
- Aug 23, 2013
- 415
Given: Simply Supported composite T-beam of length L with distributed load W. Top portion, the flange, will be attached to another member to make a composite section. Lets call A1 the flanges area and A2 the web area.
Alright here's my question. When designing composite beams, the shear to be transferred is calculated using the equation V*Q/I.
My question is, isn't this conservative? If you use a stress-strain relationship for a simply supported beam, calculate the moment of inertia, then calculate the average stress in the flange (lets call that stress F. yc is the neutral axis to the compression fiber, y1 is the center of gravity of the flange. F=M/S=W*L^2*(yc-y1)/(8*Ix)), the force is F*A1. Now this force needs to be developed from both ends to the midspan, so the maximum force must be developed by L/2.
Alright so F*A1 = M*y/I*A1 (y is the centroid of A1 so we have AVERAGE stress over the flange, the stress is trapezoidal as we are assuming the neutral axis lies outside the top flange).
Since this FORCE must be attained by midspan, the shear stress/foot would = W*L(yc-y1)*A1/(4*I)
Let's break down the V*Q/I so they're comparable. V*Q/I = W*L/2 * A1*(yc-y1) = W*L*(yc-y1)*A1/(2*I)
so V*Q/I gives twice as much force to accomodate. This (to me) seems to be because V*Q/I gives the maximum shear/foot (when V is at its max). But, for a distributed load on a simply supported beam, V is at its max at an infinitesimal point only.
Doesn't it make more sense to have enough fasteners to develop the necessary flange load by midspan instead of putting fasteners in to accommodate the maximum shear stress/foot, which only occurs over a small location?
Shouldn't it be developed for the AVERAGE shear stress? That would make them equivalent! ( the shear is triangular)
Thoughts?
Alright here's my question. When designing composite beams, the shear to be transferred is calculated using the equation V*Q/I.
My question is, isn't this conservative? If you use a stress-strain relationship for a simply supported beam, calculate the moment of inertia, then calculate the average stress in the flange (lets call that stress F. yc is the neutral axis to the compression fiber, y1 is the center of gravity of the flange. F=M/S=W*L^2*(yc-y1)/(8*Ix)), the force is F*A1. Now this force needs to be developed from both ends to the midspan, so the maximum force must be developed by L/2.
Alright so F*A1 = M*y/I*A1 (y is the centroid of A1 so we have AVERAGE stress over the flange, the stress is trapezoidal as we are assuming the neutral axis lies outside the top flange).
Since this FORCE must be attained by midspan, the shear stress/foot would = W*L(yc-y1)*A1/(4*I)
Let's break down the V*Q/I so they're comparable. V*Q/I = W*L/2 * A1*(yc-y1) = W*L*(yc-y1)*A1/(2*I)
so V*Q/I gives twice as much force to accomodate. This (to me) seems to be because V*Q/I gives the maximum shear/foot (when V is at its max). But, for a distributed load on a simply supported beam, V is at its max at an infinitesimal point only.
Doesn't it make more sense to have enough fasteners to develop the necessary flange load by midspan instead of putting fasteners in to accommodate the maximum shear stress/foot, which only occurs over a small location?
Shouldn't it be developed for the AVERAGE shear stress? That would make them equivalent! ( the shear is triangular)
Thoughts?
