Question 9.11: The intensity of loading on simple supported beam of spam 10...
The intensity of loading on simple supported beam of spam 10 m increases uniformly from 10 kN/m at the left support to 20 kN/m at the right support. Find the position and magnitude of maximum bending moment.
(U.P.T.U., Ist Sem, 2004−05)
The load lying over simple supported beam is a combination of U.D.L and U.V.L.
ΣY = 0,
R_{A}+R_{E} = Downward load of U.D.L + Downward load of U.V.L.
=(10 \times 10)+\left(\frac{1}{2} \times 10 \times(20-10)\right)
= 100 + 50 ….. (1)
R_{A}+R_{E}=150
\Sigma M _{ A }=0,
R_{B} \times 10=(10 \times 10) \times \frac{10}{2}+\left(\frac{1}{2} \times 10 \times 10\right) \times\left(\frac{2}{3} \times 10\right)
R_{B} = 83.33 kN
R_{A} = 66.67 kN
Shear Force Diagram:
(S . F .) x_{1} x_{1}^{1}=+R_{A}- Load of U.D.L up to x_{1} – Load of U.V.L up to x_{1}
Note: Here sign convention has been reversed as we are considering x_{1} x_{1}^{1} from left side i.e., reverse side from earlier questions.
Load of U.D.L. up to x_{1} = 10·x_{1}
Load of U.V.L. up to x_{1} = area of small triangle
=\frac{1}{2} \times x_{1} \times h
=\frac{1}{2} \times x_{1} \times x_{1}
= x_{1} ^{2} / 2
At 10 m distance (i.e., A to B) intensity of load = 10 kN/m
intensity of load at x_{1} distance will be given by =\frac{10 \times x_{1}}{10}
h = x_{1}
Thus equation (1) becomes,
(S F) x_{1} x_{1}^{1}=66.67-10 x_{1}-\frac{x_{1}^{2}}{2}
(S F)_{A}=+66.67 kN
(S F)_{B}=66.67-10 \times 10-\frac{(10)^{2}}{2}
\quad= −83.33 kN
Let the shear force changes its sign at distance ‘a’ from end A.
(S . F .)_{c}=66.67-10 a-\frac{a^{2}}{2}
0=66.67-10 a-\frac{a^{2}}{2}
a^{2}+20 a-133.34=0
a=\frac{-20 \pm \sqrt{(20)^{2}-4 \cdot 1 \cdot(-133.4)}}{2 \times 1}
a=\frac{-20 \pm 30.55}{2} \text { i.e. } a=-25.28 m \text { or } 5.28 m
a = 5.28 m as negative value does not lie on the beam.
Bending Moment Diagram:
(\text { B.M. }) x_{1} x_{1}^{1}=+R_{A} \times x_{1}-10 \times x_{1} \times \frac{x_{1}}{2}-\left(\frac{1}{2} \times x_{1} \times x_{1}\right) \times \frac{x_{1}}{3}
\quad=66.67 x_{1}-5 x_{1}^{2}-\frac{x_{1}^{3}}{6}
(B M)_{A}=0
(B M)_{B}=66.67 \times 10-5(10)^{2}-\frac{(10)^{3}}{6}=0
(B M)_{C}=66.67 \times 5.28-5(5.28)^{2}-\frac{(5.28)^{3}}{6}
= 188.09 kNm