A short, built-up column made up of a rough-cut $4^{\prime \prime }\times 4^{\prime \prime }$ timber member, reinforced with steel side plates, must support an axial load of 20 k. Determine the stress that develops in the timber and in the steel plates if
$E_{timber} =1.76\times 10^3 \space ksi \space ; \space E_{steel} =29\times 10^3 \space ksi$
The rigid plates at the top and bottom allow the timber and steel side plates to deform uniformly under the action of $P$ .

Question

A short, built-up column made up of a rough-cut [latex]4^{\prime \prime } imes 4^{\prime \prime }[/latex] timber member, reinforced with steel side plates, must support an axial load of 20 k. Determine the stress that develops in the timber and in the steel plates if

[latex]E_{timber} =1.76 imes 10^3 \space ksi \space ; \space E_{steel} =29 imes 10^3 \space ksi[/latex]

The rigid plates at the top and bottom allow the timber and steel side plates to deform uniformly under the action of [latex]P[/latex].

Accepted Answer

From the condition of equilibrium:

[latex]f_{t} =[/latex] Stress developed in the timber

[latex] f_{s}= [/latex] Stress developed in the steel

[latex]\left[\sum{F_{y}=0 } ight] - 20 \space k + f_{s} A_{s} +f_{t} A_{t} =0[/latex]

where

[latex]A_{s}=2 imes \frac{1}{4}^{\prime \prime } imes 4^{\prime \prime }=2 \space in.^2 [/latex]

[latex] A_{t}= 4^{\prime \prime } imes 4^{\prime \prime }=16 \space in.^2 [/latex]

[latex] 2 \space in.^2 (f_{s} )+16 \space in.^2 (f_{t})=20 \space k[/latex]

Obviously, the equation of equilibrium alone is insufficient to solve for the two unknowns. Therefore, another equation, involving a deformation relationship, must be written.

[latex]\delta =\delta _{t} = \delta _{s} [/latex]

Because both the timber and the steel members are of equal length , [latex]L_{t} = L_{s} \space ; [/latex] then ,

[latex]\varepsilon =\frac{\delta _{t}}{L_{t}} =\frac{\delta _{s} }{ L_{s} } \space ; \space \space \varepsilon_{t} = \varepsilon_{s} [/latex]

but [latex]E=\frac{f}{ \varepsilon} [/latex] or [latex]\varepsilon=\frac{f}{E}[/latex]

[latex] herefore \frac{f_{t} }{E_{t} } =\frac{f_{s} }{E_{s} } \space ; \space \space f_{t}=f_{s}\frac{E_{t}}{E_{s}} [/latex]

[latex]f_{t}=\frac{1.76 imes 10^3 \space ksi}{29 imes 10^3 \space ksi} (f_{s})=0.061 \space (f_{s})[/latex]

Substituting into the equilibrium equation,

[latex]2 \space in.^2(f_{s})+16 \space in.^2 (0.061 \space f_{s})=20 \space k[/latex]

[latex]2f_{s}+0.976f_{s}=20 \space ksi \space ; \space 2.976 \space f_{s}=20 \space ksi[/latex]

[latex]f_{s}=6.72 \space ksi[/latex]

[latex]f_{t}=0.061 (6.72 \space ksi )=0.41 \space ksi[/latex]

Question 5.20: A short, built-up column made up of a rough-cut 4^'' × 4^'' ...

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