Interaction Force between Two Beams Placed One on Top of the Other Two simply supported beams are situated one on top of the other as shown in Figure 4.10. The top beam is subjected to a uniformly distributed load of intensity w. Find: a. The interaction force R at midspan C acting upward on the

Question

Interaction Force between Two Beams Placed One on Top of the Other

Two simply supported beams are situated one on top of the other as shown in Figure 4.10. The top beam is subjected to a uniformly distributed load of intensity [latex]w[/latex].

Find:

a. The interaction force [latex]R[/latex] at midspan [latex]C[/latex] acting upward on the beam [latex]AB[/latex] and acting downward on beam [latex]DE[/latex]

b. The maximum moment and deflection in beam [latex]AB[/latex]

Assumption: The beams are supported in such a way that they are to deflect by the same amount at the junction [latex]C[/latex].

Accepted Answer

Both beams are statically indeterminate to the first degree. We select [latex]R[/latex] as redundant and treat it as an unknown load. Considering the two beams in turn and using the data in Table A.9, the deflections at the center are as follows. For beam [latex]AB[/latex], due to load [latex]u^{\prime}[/latex] and owing to [latex]R[/latex],

[latex] \upsilon _w=\frac{5 w(2 a)^4}{384 E I_a}, \quad \upsilon _R=-\frac{R(2 a)^3}{48 E I_a} [/latex]

The total downward deflection is therefore

[latex] \upsilon _a=\frac{5 w a^4}{24 E I_a}-\frac{R a^3}{6 E I_a} [/latex]

For beam [latex]DE[/latex], due to [latex]R[/latex], the downward deflection is

[latex] \upsilon _b=\frac{R b^3}{6 E I_b} [/latex]

a. Equating the two expressions for deflections [latex]\upsilon _{a} = υ_{b}[/latex] and solving for the interaction force, we have

[latex]R = \frac {1.25wa} {\alpha}[/latex] (4.21)

where

[latex] \alpha=1+\left(\frac{b}{a} ight)^3 \frac{I_a}{I_b} [/latex]

Comment: If beam DE is rigid [latex] ext { (i.e., } I_b ightarrow \infty ext { ), } [/latex] then [latex]\alpha[/latex] = 1 and [latex]R = 1.25wa[/latex], which is equal to the central reaction for the beam resting on three simple supports.

b. Maximum bending moment and deflection in beam [latex]AB[/latex] occurring at the center are, respectively,

[latex] M_C=\frac{1}{2} w a^2-\frac{1}{2} R a=\frac{1}{2} w a^2\left(1-\frac{1.25}{\alpha} ight) [/latex] (4.22a)

[latex]\upsilon _C=\frac{5 w(2 a)^4}{384 E I_a}\left(1-\frac{1}{\alpha} ight)=\frac{5 w a^4}{24 E I_a}\left(1-\frac{1}{\alpha} ight) [/latex] (4.22b)

Comments: We see from the preceding results that, as beam [latex]DE[/latex] is made stiffer by either reducing its span [latex]2b[/latex] or increasing its moment of inertia [latex]I_{b}[/latex], the value of [latex]\alpha[/latex] decreases and hence the value of [latex]R[/latex] increases. This decreases the deflection and also reduces the bending moment in the loaded beam [latex]AB[/latex].

Question 4.7: Interaction Force between Two Beams Placed One on Top of the...

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