The rigid bar DEF is welded at point D to the uniform steel beam AB. For the loading shown, determine (a) the equation of the elastic curve of the beam, (b) the deflection at the midpoint C of the beam. Use E = 29 × 10^6 psi.

Question

The rigid bar DEF is welded at point D to the uniform steel beam AB. For the loading shown, determine (a) the equation of the elastic curve of the beam, (b) the deflection at the midpoint C of the beam. Use E = 29 × 10[latex]^6[/latex] psi.

Accepted Answer

Bending Moment. The equation defining the bending moment of the beam was obtained in Sample Prob. 5.10. Using the modified loading diagram shown and expressing x in feet, we had [Eq. (3)]:

[latex]M(x)=-25 x^{2}+480 x-160\langle x-11 angle^{1}-480\langle x-11 angle^{0}[/latex] lb · ft

a. Equation of the Elastic Curve. Using Eq. (8.4), we write

[latex]t _{x y}=\frac{3}{2} \frac{P}{A}\left(1-\frac{y^{2}}{c^{2}} ight)[/latex] (8.4)

[latex]E I\left(d^{2} y / d x^{2} ight)=-25 x^{2}+480 x-160\langle x-11 angle^{1}-480\langle x-11 angle^{0} \quad lb \cdot ft[/latex] (1)

and, integrating twice in x,

[latex]E I u =-8.333 x^{3}+240 x^{2}-80\langle x-11 angle^{2}-480\langle x-11 angle^{1}+C_{1} \quad [/latex] lb · ft² (2)

[latex]\begin{aligned}E I y=-2.083 x^{4}+80 x^{3}-26.67\langle x-11 angle^{3} &-240\langle x-11 angle^{2} \&+C_{1} x+C_{2} \end{aligned}[/latex] lb · ft³ (3)

Boundary Conditions.

[x = 0, y = 0] : Using Eq. (3) and noting that each bracket [latex]\langle angle[/latex] contains a negative quantity and, thus, is equal to zero, we find [latex]C_{2}=0[/latex].

[x = 16 ft, y = 0] : Again using Eq. (3) and noting that each bracket contains a positive quantity and, thus, can be replaced by a parenthesis, we write

[latex]0=-2.083(16)^{4}+80(16)^{3}-26.67(5)^{3}-240(5)^{2}+C_{1}(16)[/latex]

[latex]C_{1}=-11.36 imes 10^{3}[/latex]

Substituting the values found for [latex]C_{1}[/latex] and [latex]C_{2}[/latex] into Eq. (3), we have

[latex]E I y=-2.083 x^{4}+80 x^{3}-26.67\langle x-11 angle^{3}-240\langle x-11 angle^{2} - 11.36 × 10^3x[/latex] lb · ft³ [latex]\left(3^{\prime} ight)[/latex]

To determine EI, we recall that E = 29 × 10[latex]^6[/latex] psi and compute

[latex]I=\frac{1}{12} b h^{3}=\frac{1}{12}(1 ext { in. })(3 ext { in. })^{3}=2.25 in ^{4}[/latex]

[latex]E I=\left(29 imes 10^{6} psi ight)\left(2.25 in ^{4} ight)=65.25 imes 10^{6} lb \cdot in ^{2}[/latex]

However, since all previous computations have been carried out with feet as the unit of length, we write

[latex]E I=\left(65.25 imes 10^{6} lb \cdot in ^{2} ight)(1 ft / 12 ext { in. })^{2}=453.1 imes 10^{3} lb \cdot ft ^{2}[/latex]

b. Deflection at Midpoint C. Making x = 8 ft in Eq.[latex]\left(3^{\prime} ight)[/latex], we write

[latex]E I y_{C}=-2.083(8)^{4}+80(8)^{3}-26.67\langle-3 angle^{3}-240\langle-3 angle^{2}-11.36 imes 10^{3}(8)[/latex]

Noting that each bracket is equal to zero and substituting for EI its numerical value, we have

[latex]\left(453.1 imes 10^{3} lb \cdot ft ^{2} ight) y_{C}=-58.45 imes 10^{3} lb \cdot ft ^{3}[/latex]

and, solving for [latex]y_{C}[/latex]: [latex]y_{C}=-0.1290 ft[/latex] [latex]y_{C}=-1.548 in.[/latex]

Note that the deflection obtained is not the maximum deflection.

Question 9.S-P.5: The rigid bar DEF is welded at point D to the uniform steel ...

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