A simply supported beam with a gradually varying load from zero at ‘B’ and ‘w’ per unit length at ‘A’ is shown in the below figure. The shear force at ‘B’ is equal to
A. $$\frac{{{\text{w}}l}}{6}$$
B. $$\frac{{{\text{w}}l}}{3}$$
C. $${\text{w}}l$$
D. $$\frac{{2{\text{w}}l}}{3}$$
Answer: Option A
Solution (By Examveda Team)
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A. Equal to
B. Less than
C. Greater than
D. None of these
A. $$\frac{{{\text{w}}l}}{6}$$
B. $$\frac{{{\text{w}}l}}{3}$$
C. $${\text{w}}l$$
D. $$\frac{{2{\text{w}}l}}{3}$$
The columns whose slenderness ratio is less than 80, are known as
A. Short columns
B. Long columns
C. Weak columns
D. Medium columns
Takr Sum of all force in y direction = 0
We get , Ra + Rb = WL / 2
Where WL / 2 is load due to UVL
Then take sum of moment @ A = 0
Then we get ,
WL^2/ 6 - Rb L = 0
Rb = WL /6
Resolve the distributed loads into point load by using the formula (1/2*b*h) ,that will give u "wl/2" and it would be L/3 from A(base) and 2L/3 from B(apex) .Get the reacting component at point A by taking the moment about point B,that will give you "wl/3",then get the share force at be by taking ∑fy=0,that will be wl/2-wl/3=wl/6.
Wl/3
Convert the UVL into a Point Load by calculating the area of the triangular loading. i.e. (1/2)*L*W.
The load will act at CG of the triangle i.e. at L/3 from support A.
Now to calculate the reaction at B, take summation of the moment at A equal to zero(Equilibrium equation).
i can't understand. please described it.