For the uniform beam and loading shown, determine the reactions at the supports.

Question

Accepted Answer

Castigliano's Theorem. The beam is indeterminate to the first degree and we choose the reaction [latex]\mathbf{R}_{A}[/latex] as redundant. Using Castigliano's theorem, we determine the deflection at [latex]A[/latex] due to the combined action of [latex]\mathbf{R}_{A}[/latex] and the distributed load. Since [latex]E I[/latex] is constant, we write

[latex]y_{A}=\int \frac{M}{E I}\left(\frac{\partial M}{\partial R_{A}} ight) d x=\frac{1}{E I} \int M \frac{\partial M}{\partial R_{A}} d x[/latex] (1)

The integration will be performed separately for portions [latex]A B[/latex] and [latex]B C[/latex] of the beam. Finally, [latex]\mathbf{R}_{A}[/latex] is obtained by setting [latex]y_{A}[/latex] equal to zero.

Free Body: Entire Beam. We express the reactions at [latex]B[/latex] and [latex]C[/latex] in terms of [latex]R_{A}[/latex] and the distributed load

[latex]R_{B}=\frac{9}{4} w L-3 R_{A} \quad R_{C}=2 R_{A}-\frac{3}{4} w L[/latex] (2)

Portion AB of Beam. Using the free-body diagram shown, we find

[latex]M_{1}=R_{A} x-\frac{w x^{2}}{2} \quad \frac{\partial M_{1}}{\partial R_{A}}=x[/latex]

Substituting into Eq. (1) and integrating from [latex]A[/latex] to [latex]B[/latex], we have

[latex]\frac{1}{E I} \int M_{1} \frac{\partial M}{\partial R_{A}} d x=\frac{1}{E I} \int_{0}^{L}\left(R_{A} x^{2}-\frac{w x^{3}}{2} ight) d x=\frac{1}{E I}\left(\frac{R_{A} L^{3}}{3}-\frac{w L^{4}}{8} ight)[/latex] (3)

Portion BC of Beam. We have

[latex]M_{2}=\left(2 R_{A}-\frac{3}{4} w L ight) v-\frac{w v^{2}}{2} \quad \frac{\partial M_{2}}{\partial R_{A}}=2 v[/latex]

Substituting into Eq. (1) and integrating from [latex]C[/latex], where [latex]v=0[/latex], to [latex]B[/latex], where [latex]v=\frac{1}{2} L[/latex], we have

[latex] \begin{aligned} \frac{1}{E I} \int M_{2} \frac{\partial M_{2}}{\partial R_{A}} d v & =\frac{1}{E I} \int_{0}^{L / 2}\left(4 R_{A} v^{2}-\frac{3}{2} w L v^{2}-w v^{3} ight) d v \ & =\frac{1}{E I}\left(\frac{R_{A} L^{3}}{6}-\frac{w L^{4}}{16}-\frac{w L^{4}}{64} ight)=\frac{1}{E I}\left(\frac{R_{A} L^{3}}{6}-\frac{5 w L^{4}}{64} ight) & (4) \end{aligned} [/latex]

Reaction at [latex]\boldsymbol{A}[/latex]. Adding the expressions obtained in (3) and (4), we determine [latex]y_{A}[/latex] and set it equal to zero

[latex]y_{A}=\frac{1}{E I}\left(\frac{R_{A} L^{3}}{3}-\frac{w L^{4}}{8} ight)+\frac{1}{E I}\left(\frac{R_{A} L^{3}}{6}-\frac{5 w L^{4}}{64} ight)=0[/latex]

Solving for [latex]R_{A}, \quad R_{A}=\frac{13}{32} w L \quad R_{A}=\frac{13}{32} w L \uparrow[/latex]

Reactions at [latex]\boldsymbol{B}[/latex] and [latex]\boldsymbol{C}[/latex]. Substituting for [latex]R_{A}[/latex] into Eqs. (2), we obtain

[latex]R_{B}=\frac{33}{32} w L \uparrow \quad R_{C}=\frac{w L}{16} \uparrow[/latex]

Question 11.SP.7: For the uniform beam and loading shown, determine the reacti...

Related Answered Questions

The cantilever beam AB supports a uniformly distributed load w (Fig.11.41). Determine the ...

Verified Answer:

Determine the reactions at the supports for the prismatic beam and loading shown ...

Verified Answer:

For the truss and loading of Sample Prob. 11.4, determine the vertical deflection of joint ...

Verified Answer:

A load P is supported at B by three rods of the same material and the same cross-sectional ...

Verified Answer:

A load P is supported at B by two rods of the same material and of the same ...

Verified Answer:

The cantilever beam AB supports a uniformly distributed load w and a concentrated load P ...

Verified Answer:

Members of the truss shown consist of sections of aluminum pipe with the cross-sectional ...

Verified Answer:

The block D of mass m is released from rest and falls a distance h before it strikes the ...

Verified Answer:

A torque T is applied at the end D of shaft BCD (Fig. 11.32). Knowing that both portions ...

Verified Answer:

Determine the deflection of end A of the cantilever beam AB (Fig. 11.31), taking into ...

Verified Answer: