Question 9.10: Determine the angle of rotation θB and deflection δB at the ......
Determine the angle of rotation θ_B and deflection δ_B at the free end B of a cantilever beam AB supporting a concentrated load P Fig. 9-24).
Note: The beam has a length L and constant flexural rigidity EI.
Use a four-step problem-solving approach.
1. Conceptualize: By inspection of the beam and its loading, the angle of rotation θ_B is clockwise and the deflection δ_B is downward (Fig. 9-24). Therefore, use absolute values when applying the moment-area theorems.
2. Categorize:
M/EI diagram: The bending-moment diagram is triangular in shape with the moment at the support equal to –PL. Since the flexural rigidity EI is constant, the M/EI diagram has the same shape as the bending-moment diagram, as shown in the last part of Fig. 9-24.
3. Analyze:
Angle of rotation: From the first moment-area theorem, the angle θ_{B/A} between the tangents at points B and A is equal to the area of the M/EI diagram between those points. This area, labeled as A_1, is determined as
\quad\quad\quad\quad A_{1}={\frac{1}{2}}(L)\left({\frac{P L}{E I}}\right)={\frac{P L^{2}}{2E I}}
This is the absolute value of the area.
The relative angle of rotation between points A and B (from the first theorem) is
\quad\quad\quad\quad \theta_{{}_{\scriptstyle B/A}}=\theta_{{}_{B}}-\theta_{{}_{A}}=A_{1}={\frac{{P}{ L}^{2}}{2E{ I}}}
The tangent to the deflection curve at support A is horizontal (θ_A = 0), so
\quad\quad\quad\quad\theta_{B}={\frac{P L^{2}}{2E I}} \quad\quad\quad\quad (9-78)
This result agrees with the formula for θ_B given in Case 4 of Table H-1, Appendix H.
Deflection: The deflection δ_B at the free end is obtained from the second moment-area theorem. In this case, the tangential deviation t_{B/A} of point B from the tangent at A is equal to the deflection δ_B itself (see Fig. 9-24). The first moment of the area of the M/EI diagram, evaluated with respect to point B, is
\quad\quad\quad\quad{{Q}}_{1}={{A}}_{1}{\overline{{x}}}={\Bigg\lgroup}{\frac{P L^{2}}{2E I}}{\Bigg\rgroup}{\Bigg\lgroup}{\frac{2L}{3}}{\Bigg\rgroup}={\frac{P L^{3}}{3E I}}
Note again that only absolute values are used here.
From the second moment-area theorem, the deflection δ_B is equal to the first moment Q_1. Therefore,
\quad\quad\quad\quad\delta_{B}=\frac{{P}{L}^{3}}{2E{I}}\quad\quad (9-79)
4. Finalize: This result also appears in Case 4 of Table H-1.
| 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) |