Question 9.14: A cantilever beam ACB having a length L and two different mo......
A cantilever beam ACB having a length L and two different moments of inertia I and 2I supports a concentrated load P at the free end A (Figs. 9-29a and b).
Determine the deflection \delta_A at the free end.
Use a four-step problem-solving approach.
1. Conceptualize: In this example, use the method of superposition to determine the deflection \delta_A at the end of the beam. Begin by recognizing that the deflection consists of two parts: the deflection due to bending of part AC of the beam and the deflection due to bending of part CB. Find these deflections separately and then superpose them to obtain the total deflection.
2. Categorize:
Deflection due to bending of part AC of the beam: Imagine that the beam is held rigidly at point C, so that the beam neither deflects nor rotates at that point (Fig. 9-29c). Now calculate the deflection \delta_1 of point A in this beam. Since the beam has length L/2 and moment of inertia I, its deflection (see Case 4 of Table H-1, Appendix H) is
\quad\quad\quad\quad {\delta}_{1}=\frac{P(L/2)^{3}}{3E I}=\frac{P L^{3}}{24E I}\quad\quad (a)
Deflection due to bending of part CB of the beam: Part CB of the beam also behaves as a cantilever beam (Fig. 9-29d) and contributes to the deflection of point A. The end of this cantilever is subjected to a concentrated load P and a moment PL/2.
Therefore, the deflection \delta_C and angle of rotation \theta_C at the free end (Fig. 9-29d) are (see Cases 4 and 6 of Table H-1):
This deflection and angle of rotation make an additional contribution δ_2 to the deflection at end A (Fig. 9-29e). Again visualize part AC as a cantilever beam, but now its support (at point C) moves downward by the amount δ_C and rotates counterclockwise
through the angle \theta_C (Fig. 9-29e). These rigid-body displacements
produce a downward displacement at end A equal to
\quad\quad\quad\quad\delta_{2}=\delta_{C}+\theta_{C}{\Bigg\lgroup}{\frac{L}{2}}{\Bigg\rgroup}={\frac{5P L^{3}}{96E I}}+{\frac{3P L^{2}}{16E I}}{\Bigg\lgroup}{\frac{L}{2}}{\Bigg\rgroup}={\frac{7P L^{3}}{48E I}}\quad\quad(b)
Total deflection: The total deflection δ_A at the free end A of the original cantilever beam (Fig. 9-29f) is equal to the sum of the deflections δ_1 ~ and ~ δ_2:
\quad\quad\quad\quad{δ}_{A} = {δ}_{1} + {δ}_{2}=\frac{P L^{3}}{24E I}+\frac{7P L^{3}}{48E I}=\frac{3P L^{3}}{16E I}\quad\quad(9-90)
4. Finalize: This example illustrates one of the many ways that the principle of superposition may be used to find beam deflections.
| Table H-1 | |
| Deflections and Slopes of Cantilever Beams | |
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Notation: v = deflection in the y direction (positive upward) v’ = dv/dx = slope of the deflection curve δ_B = -v(L) = deflection at end B of the beam (positive downward) θ_B = -v^\prime(L) = angle of rotation at end B of the beam (positive clockwise) EI = constant |
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\nu=-{\frac{q x^{2}}{24E I}}(6L^{2}-4L x+x^{2})\quad\quad v^\prime = {\frac{q x}{6E I}}(3L^{2} – 3 L x+x^{2}) \\ δ_B = {\frac{q L^4}{8E I}} \quad \theta_B = {\frac{q L^3}{6E I}} |
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\displaystyle{v=-\frac{q x^{2}}{24E I}(6a^{2}-4\alpha x+x^{2})}\qquad\qquad(0\leq x\leq a) \\ \displaystyle{v^\prime=-\frac{q x}{6E I}(3a^{2}- 3\alpha x+x^{2})}\qquad\qquad(0\leq x\leq a) \\ \displaystyle{v=-\frac{q a^{3}}{24E I}(4x – a)} \quad\quad \displaystyle{v^\prime=-\frac{q a^{3}}{6E I}}\qquad(a\leq x\leq L) \\ At \, x= a : v = -\frac{q a^{4}}{8E I} \quad v^\prime = -\frac{q a^{3}}{6E I}\\ \delta_B = {\frac{q a^{3}}{24E I}(4L – a)}\quad \theta_B = \frac{q a^{3}}{6E I} |
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\displaystyle{v=-\frac{qb x^{2}}{12E I}(6L+3\alpha – 2x)}\qquad\qquad(0\leq x\leq a) \\ \displaystyle{v^\prime= -\frac{qb x}{2E I}(L+ \alpha – x)}\qquad\qquad(0\leq x\leq a) \\ \displaystyle{v=-\frac{q }{24E I}(x^4 -4Lx^3 +6L^2x^2 – 4a^3x + a^4)} \qquad\qquad(a\leq x\leq L) \\ \displaystyle{v^\prime=-\frac{q }{6E I}(x^3 -3Lx^2 + 3L^2x – a^3)} \qquad\qquad(a\leq x\leq L)\\ AT x = a: \, v = -\frac{qa^2b}{12EI}(3L + a) \quad v^\prime = -\frac{qabL}{2EI}\\ \delta_B = {\frac{q }{24E I}(3L^4 -4 a^3L + a^4)}\quad \theta_B = \frac{q }{6E I}(L^3 – a^3) |
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v = -\frac{Px^2}{6EI}(3L – x) \quad v^\prime = -\frac{Px}{2EI}(2L – x)\\ \delta_B = {\frac{PL^3}{3E I}}\quad \theta_B = \frac{PL^2}{2E I} |
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\displaystyle{v=-\frac{Px^{2}}{6E I}(3a – x)}\quad \displaystyle{v^\prime=-\frac{Px}{2E I}(2a- x)}\qquad\qquad(0\leq x\leq a) \\ \displaystyle{v=-\frac{Pa^{2}}{6E I}(3x – a)} \quad\quad \displaystyle{v^\prime=-\frac{Pa^{2}}{2E I}}\qquad(a\leq x\leq L) \\ At \, x= a : v = -\frac{Pa^{3}}{3E I} \quad v^\prime = -\frac{Pa^{2}}{2E I}\\ \delta_B = {\frac{P a^{2}}{6E I}(3L – a)}\quad \theta_B = \frac{P a^{2}}{2E I} |
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\nu=-\frac{M_{0}x^{2}}{2E I}\quad\quad \nu^{\prime}=-\frac{M_{0}x}{E I} \\ \delta_B=-\frac{M_{0}L^{2}}{2E I}\quad\quad \theta_{B}=-\frac{M_{0}L}{E I} |
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\displaystyle{v=-\frac{M_0 x^{2}}{2E I}}\qquad \displaystyle{v^\prime=-\frac{M_0 x}{E I}}\qquad(0\leq x\leq a) \\ \displaystyle{v=-\frac{M_0 a}{2E I}}(2x – a) \quad \displaystyle{v^\prime=-\frac{M_0 a}{E I}}\qquad(0\leq x\leq a) \\ At \, x= a : \displaystyle{v=-\frac{M_0 a^{2}}{2E I}}\qquad \displaystyle{v^\prime=-\frac{M_0 a}{E I}}\\ \delta_B = {\frac{M_0 a}{2E I}(2L – a)}\quad \theta_B = \frac{M_0 a}{E I} |
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\displaystyle{v=-\frac{q_0 x^{2}}{120LE I}(10L^{3} – 10L^2 x + 5Lx^{2} – x^3)} \\ \displaystyle{v^\prime=-\frac{q_0 x}{24LE I}(4L^{3}- 6L^2 x + 4Lx^{2} – x^3)} \\ \delta_B = {\frac{q_0 L^{4}}{30E I}}\quad \theta_B = \frac{q_0 L^{3}}{24E I} |
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\displaystyle{v=-\frac{q_0 x^{2}}{120LE I}(20L^{3} – 10L^2 x + x^3)} \\ \displaystyle{v^\prime=-\frac{q_0 x}{24LE I}(8L^{3}- 6L^2 x + x^3)} \\ \delta_B = {\frac{11q_0 L^{4}}{120E I}}\quad \theta_B = \frac{q_0 L^{3}}{8E I} |
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\displaystyle{v=-\frac{q_0 L}{3\pi ^4E I}(48L^{3}\cos {\frac{\pi x}{2L}} – 48L^3 + 3\pi ^3Lx^2 – \pi^3x^3)} \\ \displaystyle{v^\prime=-\frac{q_0 L}{\pi ^3E I}( 2\pi ^2Lx – \pi^2x^2 – 8L^{2}\sin {\frac{\pi x}{2L}})} \\ \delta_B = {\frac{2q_0 L^{4}}{3\pi^4E I}}(\pi^3 – 24)\quad \theta_B = \frac{q_0 L^{3}}{\pi^3E I}(\pi^2 – 8) |