Consider a vertically mounted, axially loaded member subject to its own distributed weight w N/m (see Fig. 4.5a). Assume that the member has a constant cross-sectional area A and a constant elastic modulus E. The total weight of the member of length L is thus W=wL. Find the displacement ux

Question

Consider a vertically mounted, axially loaded member subject to its own distributed weight w N/m (see Fig. 4.5a). Assume that the member has a constant cross-sectional area A and a constant elastic modulus E. The total weight of the member of length L is thus W=wL. Find the displacement [latex]u_{x}[/latex] at the free end [latex][i.e., δ\equiv u_{x} (x=L)].[/latex]FIGURE 4.5 A vertically loaded member subject to its own weight, given as w (force per unit length) and thus a total weightW=wL, which acts at the center of gravity. Shown is the physical problem, a free-body diagram of the whole to isolate reaction [latex]R_{x}[/latex] at the fixed support, and a free-body diagram of a part to isolate the internal force f(x).

Accepted Answer

First, let us construct a free-body diagram of the whole structure and ensure equilibrium to find the reactions (Fig. 4.5b):

[latex] \sum{F_{x}}=0,-R_{x}+W=0 ightarrow R_{x}=W=wL ,[/latex]

[latex] \sum{F_{y}}=0,R_{y}=0,[/latex]

[latex] \sum{M_{z}}=0,M_{wall}-0.[/latex]

Next, construct a free-body diagram of the parts (Fig. 4.5c) recalling that if a structure is in equilibrium, then each of its parts is in equilibrium. The force f(x) due to the weight of the member is w(L-x) at any cross section cut at a distance x from the support; at [latex]x=0, f(0)=R_{x}=wL,[/latex] the entire weight, as it should. Alternatively, in terms of the total weight of the member, the force becomes W(1-x/L) and thus

[latex] \sum{F_{x}}=0 ightarrow \int{\sigma _{xx}dA}-f=0 ightarrow \sigma _{xx}=\frac{f}{A}=\frac{W}{A}(1-x/L).[/latex]

Note that the stress is largest at x=0, where all of the weight must be borne by the material, and the stress is zero at the free end, which is free of applied loads (i.e., traction-free). Given the stress, the strain and the axial displacement can now be computed using Hooke’s law and Eq. (4.5); namely

[latex]u_{x}(x=c)-u_{x}(x=a)=\int_{a}^{c}{\frac{f(x)}{A(x)E(x)}dx },[/latex] (4.5)

[latex] \varepsilon _{xx}=\frac{1}{E}\left[\sigma _{xx}- u (\sigma _{yy}+\sigma _{zz}) ight][/latex]

with [latex]σ_{yy}[/latex] and [latex]σ_{zz}[/latex] each zero. Thus,

[latex] \varepsilon _{xx}=\frac{\sigma _{xx}}{E} ightarrow \int_{0}^{L}{\varepsilon _{xx}dx}=u_{x}(x=L)-u_{x}(x=0)=\int_{0}^{L}{\frac{W(1-x/L)}{AE}dx },[/latex]

where [latex]u_{x}=0[/latex] at [latex]x=0[/latex] (a displacement boundary condition) and the end dis-placement is

[latex] \delta =u_{x}(x=L)=\frac{W}{AE}\left(x-\frac{x^{2}}{2L}\mid ^{L}_{0} ight)=\frac{WL}{2AE}.[/latex]

Of course, the displacement at any value of x is found by integrating from 0 to x rather than from 0 to L.

Question 4.1: Consider a vertically mounted, axially loaded member subject...

Related Answered Questions