Question 9.11: Find the angle of rotation θB and deflection δB at the free ......
Find the angle of rotation θ_B and deflection \delta_B at the free end B of a cantilever beam ACB supporting a uniform load of intensity q acting over the right-hand half of the beam (Fig. 9-25). Note: The beam has a length L and constant flexural rigidity EI.
Use a four-step problem-solving approach.
1. Conceptualize: The deflection and angle of rotation at end B of the beam have the directions shown in Fig. 9-25b. These directions are known in advance, so write the moment-area expressions using only absolute values.
2. Categorize:
M/EI diagram: The bending-moment diagram consists of a parabolic curve in the region of the uniform load and a straight line in the left-hand half of the beam. Since EI is constant, the M/EI diagram has the same shape (see Fig. 9-25c).
The values of M/EI at points A and C are -3qL^2 / 8EI ~ and ~ -qL^2 / 8EI , respectively.
3. Analyze:
Angle of rotation: For the purpose of evaluating the area of the M/EI diagram, it is convenient to divide the diagram into three parts: a parabolic spandrel of area A_1, a rectangle of area A_2, and a triangle of area A_3. These areas are
\quad\quad A_{1}=\frac{1}{3}{\Bigg\lgroup}\frac{L}{2}{\Bigg\rgroup}{\Bigg\lgroup}\frac{q L^{2}}{8E I}{\Bigg\rgroup}=\frac{q L^{3}}{48E I}\quad A_{2}=\frac{L}{2}{\Bigg\lgroup}\frac{q L^{2}}{8E I}{\Bigg\rgroup}=\frac{q L^{3}}{16E I} \\ A_{3}=\frac{1}{2}{\Bigg\lgroup}\frac{L}{2}{\Bigg\rgroup}{\Bigg\lgroup}\frac{3q L^{2}}{8E I} – \frac{q L^{2}}{8E I}{\Bigg\rgroup}=\frac{q L^{3}}{16E I}
According to the first moment-area theorem, the angle between the tangents at points A and B is equal to the area of the M/EI diagram between those points.
Since the angle at A is zero, it follows that the angle of rotation θ_B is equal to the area of the diagram; thus,
\quad\quad \theta_{B}=A_{1} + A_{2} + A_{3}={\frac{7q L^{3}}{48E I}}\quad\quad(9-80)
Deflection: The deflection \delta_B is the tangential deviation of point B with respect to a tangent at point A (Fig. 9-25b). Therefore, from the second moment-area theorem, \delta_B is equal to the first moment of the M/EI diagram and is evaluated with respect to point B:
\delta_{B}={ A}_{1}\overline{x}_{1}+{ A}_{2}\overline{x}_{2}+{A}_{3}\overline{x}_{3}\quad\quad(a)
in which \overline{x_1}, \overline{x_2},~ and ~ \overline{x_3} are the distances from point B to the centroids of the respective areas. These distances are
{\overline{{x}}}_{1}={\frac{3}{4}}{\Bigg\lgroup}{\frac{L}{2}}{\Bigg\rgroup}={\frac{3L}{8}}\quad{\overline{{x}}_{2}}={\frac{L}{2}}+{\frac{L}{4}}={\frac{3L}{4}}\quad{\overline{{x}}}_{3}={\frac{L}{2}}+{\frac{2}{3}}{\Bigg\lgroup}{\frac{L}{2}}{\Bigg\rgroup}={\frac{5L}{6}}
Substitute into Eq. (a) to find
\delta_{E}\,=\,{\frac{q L^{3}}{48E I}}{\Bigg\lgroup}{\frac{3L}{8}}{\Bigg\rgroup}\,+\,{\frac{q L^{3}}{16E I}}{\Bigg\lgroup}{\frac{3L}{4}}{\Bigg\rgroup}\,+\,{\frac{q L^{3}}{16E I}}{\Bigg\lgroup}{\frac{5L}{6}}{\Bigg\rgroup}\,=\,{\frac{41q L^{4}}{384E I}}\quad\quad (9-81)
4. Finalize: This example illustrates how the area and first moment of a complex M/EI diagram can be determined by dividing the area into parts having known properties.
The results of this analysis [Eqs. (9-80) and (9-81)] can be verified by using the formulas of Case 3, Table H-1, Appendix H, and substituting a = b = L / 2.
| 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'(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' = {\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'=-\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'=-\frac{q a^{3}}{6E I}}\qquad(a\leq x\leq L) \\ At \, x= a : v = -\frac{q a^{4}}{8E I} \quad v' = -\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'= -\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'=-\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' = -\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' = -\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'=-\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'=-\frac{Pa^{2}}{2E I}}\qquad(a\leq x\leq L) \\ At \, x= a : v = -\frac{Pa^{3}}{3E I} \quad v' = -\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_{o}x^{2}}{2E I}\quad\quad \nu^{\prime}=-\frac{M_{o}x}{E I} \\ \delta_B=-\frac{M_{o}L^{2}}{2E I}\quad\quad \theta_{B}=-\frac{M_{o}L}{E I} |
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\displaystyle{v=-\frac{M_o x^{2}}{2E I}}\qquad \displaystyle{v'=-\frac{M_o x}{E I}}\qquad(0\leq x\leq a) \\ \displaystyle{v=-\frac{M_o a}{2E I}}(2x – a) \quad \displaystyle{v'=-\frac{M_o a}{E I}}\qquad(0\leq x\leq a) \\ At \, x= a : \displaystyle{v=-\frac{M_o a^{2}}{2E I}}\qquad \displaystyle{v'=-\frac{M_o a}{E I}}\\ \delta_B = {\frac{M_o a}{2E I}(2L – a)}\quad \theta_B = \frac{M_o a}{E I} |
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\displaystyle{v=-\frac{q_o x^{2}}{120LE I}(10L^{3} – 10L^2 x + 5Lx^{2} – x^3)} \\ \displaystyle{v'=-\frac{q_o x}{24LE I}(4L^{3}- 6L^2 x + 4Lx^{2} – x^3)} \\ \delta_B = {\frac{q_o L^{4}}{30E I}}\quad \theta_B = \frac{q_o L^{3}}{24E I} |
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\displaystyle{v=-\frac{q_o x^{2}}{120LE I}(20L^{3} – 10L^2 x + x^3)} \\ \displaystyle{v'=-\frac{q_o x}{24LE I}(8L^{3}- 6L^2 x + x^3)} \\ \delta_B = {\frac{11q_o L^{4}}{120E I}}\quad \theta_B = \frac{q_o L^{3}}{8E I} |
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\displaystyle{v=-\frac{q_o L}{3\pi ^4E I}(48L^{3}\cos {\frac{\pi x}{2L}} – 48L^3 + 3\pi ^3Lx^2 – \pi^3x^3)} \\ \displaystyle{v'=-\frac{q_o L}{\pi ^3E I}( 2\pi ^2Lx – \pi^2x^2 – 8L^{2}\sin {\frac{\pi x}{2L}})} \\ \delta_B = {\frac{2q_o L^{4}}{3\pi^4E I}}(\pi^3 – 24)\quad \theta_B = \frac{q_o L^{3}}{\pi^3E I}(\pi^2 – 8) |