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

Both beams are statically indeterminate to the first degree. We select R 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 AB, due to load u^{\prime} and owing to R,

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

The total downward deflection is therefore

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

For beam DE, due to R, the downward deflection is

\upsilon _b=\frac{R b^3}{6 E I_b}

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

R = \frac {1.25wa} {\alpha}     (4.21)

where

\alpha=1+\left(\frac{b}{a}\right)^3 \frac{I_a}{I_b}

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

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

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}\right)             (4.22a)

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

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