Question 9.15: A simple beam AB of a length L supports a uniform load of an......
A simple beam AB of a length L supports a uniform load of an intensity q (Fig. 9-33). (a) Evaluate the strain energy of the beam from the bending moment in the beam. (b) Evaluate the strain energy of the beam from the equation of the deflection curve. Note: The beam has constant flexural rigidity EI.
Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
Part (a): Strain energy from the bending moment.
1, 2. Conceptualize, Categorize: The reaction of the beam at support A is qL/2, so the expression for the bending moment in the beam is
\quad\quad\quad\quad M={\frac{q L x}{2}}-{\frac{q x^{2}}{2}}={\frac{q}{2}}(L x-x^{2})\quad\qquad(a)
3. Analyze: The strain energy of the beam [from Eq. (9-95a)] is
U=\int\frac{M^{2}d x}{2E I} \qquad U\;\;=\;\int\frac{E I}{2}{\Bigg\lgroup}\frac{d^{2}\nu}{d x^{2}}{\Bigg\rgroup}^{2}d x\qquad (9-95a,b)\\
\quad\quad\quad\quad U=\int_{0}^{L}\frac{M^{2}d x}{2E I}=\frac{1}{2E I}\int_{0}^{L}\left[\frac{q}{2}(L x-x^{2})\right]^{2}d x=\frac{q^{2}}{8E I}\int_{0}^{L}\left(L^{2}x^{2}\,-2L x^{2}\,+x^{4}\right)d x\quad\quad (b)
which leads to
\quad\quad\quad\quad U={\frac{q^{2}{\boldsymbol{L}}^{5}}{240E I}}\qquad\qquad\qquad(9-98)
Note that the load q appears to the second power, which is consistent with the fact that strain energy is always positive. Furthermore, Eq. (9-98) shows that strain energy is not a linear function of the loads, even though the beam itself behaves in a linearly elastic manner.
Part (b): Strain energy from the deflection curve.
1, 2. Conceptualize, Categorize: The equation of the deflection curve for a simple beam with a uniform load is given in Case 1 of Table H-2, Appendix H, as
\quad\quad\quad\quad \nu=-\frac{q x}{24E I}(L^{3}-2L x^{2}+x^{3})\qquad\quad(c)
3. Analyze: Taking two derivatives of this equation gives
\quad\quad\quad\quad {\frac{d\nu}{d x}}=-{\frac{q}{24E I}}(L^{3}-6L x^{2}+4x^{3})\quad{\frac{d^{2}\nu}{d x^{2}}}={\frac{q}{2E I}}(L x-x^{2})
Substitute the latter expression into the equation for strain energy [Eq. (9-95b)] to obtain
\quad\quad\quad\quad U\;\;=\;\int_{0}^{L}\frac{E I}{2}{\Bigg\lgroup}\frac{d^{2}\nu}{d x^{2}}{\Bigg\rgroup}^{2}d x=\frac{E I}{2}\;\int_{0}^{L}\left[\frac{q}{2E I}(L x-x^{2})\right]^{2}d x
\quad\quad\quad\quad =\frac{q^{2}}{8E I}\,\int_{0}^{L}({L}^{2}x^{2}-2L x^{3}+x^{4})d{x}\quad\quad\quad(d)
4. Finalize: The final integral in this equation is the same as the final integral in Eq. (b) which leads to the same result as before [Eq. (9-98)].
| Table H-2 | |
| Deflections and Slopes of Simple Beams | |
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Notation: v = deflection in the y direction (positive upward) v’ = dv/dx = slope of the deflection curve δ_C = -v(L/2) = deflection at end B of the beam (positive downward) x_1 = distance from support A to point of maximum deflection \delta_{max} = -v_{max} = maximum deflection (positive downward) \theta_A = -v^\prime(0) = angle of rotation at left-hand end of the beam (positive clockwise) \theta_B = v^\prime(L) = angle of rotation at right-hand end of the beam (positive counterclockwise) EI = constant |
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\begin{array}{l}{{\nu=-{\frac{q x}{24E I}}(L^{3}-2L x^{2}+x^{3})}}\\ {{\nu^\prime=-{\frac{q}{24E I}}(L^{3}-6L x^{2}+4x^{3})}}\\ {{\mathrm{~\delta_{C}= \delta_{max}= -\frac{5qL^4}{384E I} \quad \theta_A = \theta_B = \frac{qL^3}{24EI}}}}\end{array} |
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\begin{array}{l}{{\nu=-{\frac{q x}{384E I}}(9L^{3} – 24L x^{2} + 16x^{3})}} \quad\quad \left(0\leq x\leq{\frac{L}{2}}\right) \\ {{\nu^\prime=-{\frac{q}{384E I}}(9L^{3} – 72L x^{2} + 64x^{3})}}\quad\quad \left(0\leq x\leq{\frac{L}{2}}\right) \\ \nu=-{\frac{q L}{384E I}}(8x^{3} – 24L x^{2} + 17L^{2}x – L^3) \quad\quad \left(\frac{L}{2}\leq x\leq {L}\right) \\ \nu^\prime=-{\frac{q L}{384E I}}(24x^{2} – 48L x + 17L^{2}) \quad\quad \left(\frac{L}{2}\leq x\leq {L}\right) \\ {{\mathrm{~\delta_{C}= -\frac{5qL^4}{768E I} \, \theta_A = \frac{3qL^3}{128EI} \, \theta_B = \frac{7qL^3}{384EI}}}} \end{array} |
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\begin{array}{l}{{\nu=-{\frac{q x}{24LE I}}(a^{4} – 4a^3L +4a^2L^2 + 2a^2x^2 – 4aLx^2 + Lx^{3})}} \quad\quad (0 \leq x \leq a)\\ {{\nu^\prime=-{\frac{q}{24LE I}}(a^{4} – 4a^3L +6a^2L^2 + 6a^2x^2 – 12aLx^2 – 4Lx^{3})}} \quad\quad (0 \leq x \leq a) \\ \nu=-{\frac{q a^2}{24LE I}}(-a^{2}L + 4L^{2}x + a^2 x -6 Lx^2 + 2x^3) \quad\quad (a \leq x \leq {L})\\ \nu^\prime=-{\frac{q A^2}{24LE I}}(4L^{2} + a^2 – 12Lx + 6x^2) \quad\quad\quad\quad (a \leq x \leq {L}) \\ {{\mathrm{~ \theta_A = \frac{qa^2}{24LEI}(2L – a)^2 \, \theta_B = \frac{qa^2}{24LEI}(2L^2 – a^2)}}} \end{array} |
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v=-{\frac{P x}{48E I}}(3L^{2}-4x^{2})\;\;\;\;\;v^{\prime}=-{\frac{P}{16E I}}(L^{2}-4x^{2})\;\;\;\;\left(0\leq x\leq{\frac{L}{2}}\right) \\ \delta_C = \delta_{max} = {\frac{PL^3}{48E I}}\quad\quad \theta_A = \theta_B = {\frac{PL^2}{16E I}} |
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\nu=-{\frac{P b x}{6L E I}}(L^{2}\,-\,b^{2}\,-\,x^{2}\,)\quad\nu^{\prime}=-{\frac{P b}{6L E I}}(L^{2}\,-\,b^{2}\,-\,3x^{2}\,)\quad(0\leq\,x\leq a) \\ \theta_A = \frac{Pab(L + a)}{6LEI} \quad \theta_B = \frac{Pab(L + a)}{6LEI} \\ If \, a ≥ b,\quad \delta_C = \frac{Pb(3L^2 – 4b^2)}{48EI} \quad If \, a \leq b,\quad \delta_C = \frac{Pa(3L^2 – 4a^2)}{48EI} \\ If \, a ≥ b,\quad x_1 = \sqrt{\frac{L^2 – b^2}{3}} \quad\quad and \quad \delta_{max}= \frac{Pb(L^2 – b^2)^{3/2}}{9\sqrt{3}LEI} |
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v=-{\frac{P x}{6E I}}(3aL – 3a^{2} – x^2)\;\;\;\;\;v^{\prime}=-{\frac{P}{2E I}}(aL – a^{2} – x^{2})\;\;\;\;\left(0\leq x\leq a\right) \\ v=-{\frac{P a}{6E I}}(3Lx – 3x^{2} – a^2)\;\;\;\;\;v^{\prime}=-{\frac{Pa}{2E I}}(L – 2 x)\;\;\;\;\left(a\leq x\leq L – a\right) \\ \delta_C = \delta_{max} = {\frac{Pa}{24E I}}(3L^2 – 4a^2)\quad\quad \theta_A = \theta_B = {\frac{P a (L- a)}{2E I}} |
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\nu=-\frac{M_{0}x}{6 LE I}(2L^{2} – 3L x + x^{2}) \quad {\nu}^{\prime}=-\frac{M_{0}}{6L E I}(2L^{2}-6L x+3x^{2}) \\ \delta_C = \frac{M_{0}L^2}{16 E I} \quad \theta_A = \frac{M_oL}{3EI} \quad \theta_B = \frac{M_o L}{6EI} \\ x_1 = L {{\Bigg\lgroup}}1 – \frac{\sqrt{3}}{3} {{\Bigg\rgroup}} \quad and \quad \delta_{max} = \frac{M_o L^2}{9\sqrt{3}EI} |
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\nu=-\frac{M_{0}x}{24 LE I}(L^{2} – 4x^{2}) \quad {\nu}^{\prime}=-\frac{M_{0}}{24L E I}(L^{2}-12x^{2}) \quad \left(0 \leq x \leq \frac{L}{2} \right) \\ \delta_C = 0 \quad \theta_A = \frac{M_oL}{24EI} \quad \theta_B = \frac{M_o L}{24EI} |
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\displaystyle{v=-\frac{M_o x}{6LE I}}(6aL – 3a^2 – 2 L^2 – x^2)\qquad(0\leq x\leq a) \\ \displaystyle{v^\prime=-\frac{M_o }{6LE I}}(6aL – 3a^2 – 2 L^2 – 3x^2)\qquad(0\leq x\leq a) \\ At \, x= a : \displaystyle{v=-\frac{M_o ab}{3LE I}}(2a – L)\qquad \displaystyle{v^\prime=-\frac{M_o }{3LE I}}(3aL – 3a^2 – L^2)\\ \theta_A = {\frac{M_o }{6LE I}(6 a L – 3a^2 – 2L^2)}\quad \theta_B = \frac{M_o }{6LE I}(3a^2 – L^2) |
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\nu=-\frac{M_{0}x}{2E I}(L – x ) \quad {\nu}^{\prime}=-\frac{M_{0}}{2 E I}(L – 2 x) \\ \delta_C = \delta_{max} = \frac{M_{0}L^2}{8 E I} \quad \theta_A = \theta_B = \frac{M_o L}{2EI} |
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\displaystyle{v=-\frac{q_o x}{360LE I}(7L^{4} – 10L^2 x^2 + 3Lx^{4} )} \\ \displaystyle{v^\prime=-\frac{q_o }{360LE I}(7L^{4} – 30L^2 x^2 + 15x^{4} )} \\ \delta_C = {\frac{5q_o L^{4}}{768E I}}\quad \theta_A = \frac{7q_o L^{3}}{360E I} \quad \theta_B = \frac{q_o L^{3}}{45E I} \\ x_1 = 0.5193 L \quad \delta_{max} = 0.00652\frac{q_oL^4}{EI} |
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\displaystyle{v=-\frac{q_o x}{960LE I}(5L^2 – 4x^2)^2} \quad\quad \left( 0 \leq x \leq \frac{L}{2}\right) \\ \displaystyle{v^\prime=-\frac{q_o}{192LE I}(5L^2 – 4x^2)(L^2 – 4x^2)} \quad\quad \left( 0 \leq x \leq \frac{L}{2}\right) \\ \delta_C = \delta_{max} = {\frac{q_o L^{4}}{120E I}} \quad \theta_A = \theta_B = \frac{5q_o L^{3}}{192E I} |
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\displaystyle{v=-\frac{q_o L^4}{\pi ^4E I}\sin {\frac{\pi x}{L}}} \quad \displaystyle{v^\prime=-\frac{q_o L^3}{\pi ^3E I}\cos {\frac{\pi x}{L}}} \\ \delta_C = \delta_{max} = {\frac{q_o L^{4}}{\pi^4E I}} \quad \theta_A = \theta_B = \frac{q_o L^{3}}{\pi^3E I} |