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Apply a free-body analysis to the bar BDE to find the forces exerted by links AB and DC. Evaluate the deformation of links AB and DC or the displacements of B and D. Work out the geometry to find the deflection at E given the deflections at B and D. Example 5 (cont’d) SOLUTION: Free body: Bar BDE. ∑ MB = 0.
$\delta = \dfrac{PL}{AE} = \dfrac{\sigma L}{E}$ To use this formula, the load must be axial, the bar must have a uniform cross-sectional area, and the stress must not exceed the proportional limit. If however, the cross-sectional area is not uniform, the axial deformation can be determined by considering a differential length and applying ...
> r := proc (x) r1 + (r2-r1)*(x/L) end; > A := proc (r) Pi*(r(x))^2 end; > Iz := proc (r) Pi*(r(x))^4 /4 end; > Jp := proc (r) Pi*(r(x))^4 /2 end; where r(x) is the radius, A(r) is the section area, Iz is the rectangular moment of inertia, and Jp is the polar moment of inertia.
Displacement diagrams are effectively plotting the displacement vectors of the joints as defined by the end of the bars. The displacement vector for the end of a bar is made up of two components: (1) an extension, of a magnitude defined by the bar force and the constitutive behavior of the bar which is parallel to the direction of the bar and (2) a
Consider the beam of Fig. 1.14 axially loaded along the x axis in com-pression. If a small load or displacement is applied laterally at the location of the axial load, the beam bends slightly. If the lateral load is removed, the beam returns to its straight position.
This is the slope of the deflected neutral axis as a function of x, at least within the domain 0<x<L/4. Integrating once more produces an expression for the dis-placement of the neutral axis and, again, a constant of integration. 3P vx()= – -----⋅(x 3 ⁄6)+ C 1 ⋅x + C2 EI
You use the formula displacement = PL/AE where P=load, L=length of member, A=cross-sectional area tangent to the load, and E=Young's modulus. You get: 500(*1.5m^1/3)^1/3 N/m x 1.5m = 0.00175m^2. Mar 11, 2014
