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

Question

The cantilever beam [latex]A B[/latex] supports a uniformly distributed load [latex]w[/latex] and a concentrated load [latex]\mathbf{P}[/latex] as shown (Fig. 11.40). Knowing that [latex]L=2 \mathrm{~m}[/latex], [latex]w=4 \mathrm{kN} / \mathrm{m}, P=6 \mathrm{kN}[/latex], and [latex]E I=5 \mathrm{MN} \cdot \mathrm{m}^{2}[/latex], determine the deflection at [latex]A[/latex].

Accepted Answer

The deflection [latex]y_{A}[/latex] of the point [latex]A[/latex] where the load [latex]\mathbf{P}[/latex] is applied is obtained from Eq. (11.70).

[latex]x_{j}={\frac{\partial U}{\partial P_{j}}}=\;\int_{0}^{L}{\frac{M}{E I}}{\frac{\partial M}{\partial P_{j}}}\,d x[/latex] (11.70)

Since [latex]\mathbf{P}[/latex] is vertical and directed downward, [latex]y_{A}[/latex] represents a vertical deflection and is positive downward. We have

[latex]y_{A}=\frac{\partial U}{\partial P}=\int_{0}^{L} \frac{M}{E I} \frac{\partial M}{\partial P} d x[/latex] (11.73)

The bending moment [latex]M[/latex] at a distance [latex]x[/latex] from [latex]A[/latex] is

[latex]M=-\left(P x+\frac{1}{2} w x^{2} ight)[/latex] (11.74)

and its derivative with respect to [latex]P[/latex] is

[latex]\frac{\partial M}{\partial P}=-x[/latex]

Substituting for [latex]M[/latex] and [latex]\partial M / \partial P[/latex] into Eq. (11.73), we write

[latex] \begin{gathered} y_{A}=\frac{1}{E I} \int_{0}^{L}\left(P x^{2}+\frac{1}{2} w x^{3} ight) d x \ y_{A}=\frac{1}{E I}\left(\frac{P L^{3}}{3}+\frac{w L^{4}}{8} ight) & (11.75) \end{gathered} [/latex]

Substituting the given data, we have

[latex] \begin{gathered} y_{A}=\frac{1}{5 imes 10^{6} \mathrm{~N} \cdot \mathrm{m}^{2}}\left[\frac{\left(6 imes 10^{3} \mathrm{~N} ight)(2 \mathrm{~m})^{3}}{3}+\frac{\left(4 imes 10^{3} \mathrm{~N} / \mathrm{m} ight)(2 \mathrm{~m})^{4}}{8} ight] \ y_{A}=4.8 imes 10^{-3} \mathrm{~m} \quad y_{A}=4.8 \mathrm{~mm} \downarrow \end{gathered} [/latex]

We note that the computation of the partial derivative [latex]\partial M / \partial P[/latex] could not have been carried out if the numerical value of [latex]P[/latex] had been substituted for [latex]P[/latex] in the expression (11.74) for the bending moment.

Question 11.12: The cantilever beam AB supports a uniformly distributed load...

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

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:

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: