Question 9.S-P.5: The rigid bar DEF is welded at point D to the uniform steel ...
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.
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)]:
M(x)=-25 x^{2}+480 x-160\langle x-11\rangle^{1}-480\langle x-11\rangle^{0} lb · ft
a. Equation of the Elastic Curve. Using Eq. (8.4), we write
t _{x y}=\frac{3}{2} \frac{P}{A}\left(1-\frac{y^{2}}{c^{2}}\right) (8.4)
E I\left(d^{2} y / d x^{2}\right)=-25 x^{2}+480 x-160\langle x-11\rangle^{1}-480\langle x-11\rangle^{0} \quad lb \cdot ft (1)
and, integrating twice in x,
E I u =-8.333 x^{3}+240 x^{2}-80\langle x-11\rangle^{2}-480\langle x-11\rangle^{1}+C_{1} \quad lb · ft² (2)
\begin{aligned}E I y=-2.083 x^{4}+80 x^{3}-26.67\langle x-11\rangle^{3} &-240\langle x-11\rangle^{2} \\&+C_{1} x+C_{2} \end{aligned} lb · ft³ (3)
Boundary Conditions.
[x = 0, y = 0] : Using Eq. (3) and noting that each bracket \langle \rangle contains a negative quantity and, thus, is equal to zero, we find C_{2}=0.
[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
0=-2.083(16)^{4}+80(16)^{3}-26.67(5)^{3}-240(5)^{2}+C_{1}(16)C_{1}=-11.36 \times 10^{3}
Substituting the values found for C_{1} and C_{2} into Eq. (3), we have
E I y=-2.083 x^{4}+80 x^{3}-26.67\langle x-11\rangle^{3}-240\langle x-11\rangle^{2} – 11.36 × 10^3x lb · ft³ \left(3^{\prime}\right)
To determine EI, we recall that E = 29 × 10^6 psi and compute
I=\frac{1}{12} b h^{3}=\frac{1}{12}(1 \text { in. })(3 \text { in. })^{3}=2.25 in ^{4}E I=\left(29 \times 10^{6} psi \right)\left(2.25 in ^{4}\right)=65.25 \times 10^{6} lb \cdot in ^{2}
However, since all previous computations have been carried out with feet as the unit of length, we write
E I=\left(65.25 \times 10^{6} lb \cdot in ^{2}\right)(1 ft / 12 \text { in. })^{2}=453.1 \times 10^{3} lb \cdot ft ^{2}
b. Deflection at Midpoint C. Making x = 8 ft in Eq.\left(3^{\prime}\right), we write
E I y_{C}=-2.083(8)^{4}+80(8)^{3}-26.67\langle-3\rangle^{3}-240\langle-3\rangle^{2}-11.36 \times 10^{3}(8)
Noting that each bracket is equal to zero and substituting for EI its numerical value, we have
\left(453.1 \times 10^{3} lb \cdot ft ^{2}\right) y_{C}=-58.45 \times 10^{3} lb \cdot ft ^{3}
and, solving for y_{C}: y_{C}=-0.1290 ft y_{C}=-1.548 in.
Note that the deflection obtained is not the maximum deflection.
