Question 16.11: Determine the position and magnitude of the maximum deflecti......
Determine the position and magnitude of the maximum deflection of the simply supported beam shown in Fig. P.16.11 in terms of its flexural rigidity EI.
Answer: 38.8/EI m downward at 2.9 m from left-hand support.
The uniformly distributed load is extended from D to F and an upward uniformly distributed load of the same intensity applied over DF so that the overall loading is unchanged (see Fig. S.16.11).
The support reaction at A is given by
Then
R_{\mathrm{A}}=7.7\mathrm{kN}Using Macauley’s method, the bending moment in the bay DF is
M=-7.7z+6[z-1]+4[z-3]+{\frac{1[z-3]^{2}}{2}}-{\frac{1[z-5]^{2}}{2}}Substituting in Eqs (16.31),
u^{\prime\prime}=-{\frac{M_{y}}{E I_{y y}}},\quad ν^{\prime\prime}=-{\frac{M_{x}}{E I_{x x}}} (16.31)
E I\left({\frac{\mathrm{d}^{2}v}{\mathrm{d}z^{2}}}\right)=7.7z-6[z-1]-4[z-3]-{\frac{[z-3]^{2}}{2}}+{\frac{[z-5]^{2}}{2}}
E I\left({\frac{\mathrm{d}v}{\mathrm{d}z}}\right)={\frac{7.7z^{2}}{2}}-3[z-1]^{2}-2[z-3]^{2}-{\frac{[z-3]^{3}}{6}}+{\frac{[z-5]^{3}}{6}}+C_{1}
E I v={\frac{7.7_{z}{}^{3}}{6}}-\left[z-1\right]^{3}-{\frac{2\left[z-3\right]^{3}}{3}}-{\frac{\left[z-3\right]^{4}}{24}}+{\frac{\left[z-5\right]^{4}}{24}}+C_{1}z+C_{2}
When z = 0, υ=0 so that C_{2}=0. Also when z = 6 m, υ=0. Then
0={\frac{7.7\times6^{3}}{6}}-5^{3}-{\frac{2\times3^{3}}{3}}-{\frac{3^{4}}{24}}+{\frac{1^{4}}{24}}+6C_{1}which gives
C_{1}=-21.8Guess that the maximum deflection lies between B and C. If this is the case, the slope of the beam will change sign from B to C.
At B,
E I{\biggl(}{\frac{\mathrm{d}v}{\mathrm{d}z}}{\biggr)}={\frac{7.7\times1^{2}}{2}}-21.8 which is clearly negative
At C,
E I\left({\frac{\mathrm{d}v}{\mathrm{d}z}}\right)={\frac{7.7\times3^{2}}{2}}-3\times2^{2}-21.8=+0.85The maximum deflection therefore occurs between B and C at a section of the beam where the slope is zero.
i.e.,
Simplifying,
z^{2}+7.06z-29.2=0Solving,
z=2.9m
The maximum deflection is then given by
E I v_{\mathrm{max}}={\frac{7.7\times2.9^{3}}{6}}-1.9^{3}-21.8\times2.9=-38.8i.e.,
v_{\mathrm{max}}={\frac{-38.8}{E I}}({\mathrm{downward}})