In Fig. 1.35, a uniform bar of length l m is supported at its end and loaded by a weight W kgf at its middle. In equilibrium, find the reactions $R_{1}$ and $R_{2}$ at the ends.

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Solution

$R_{1} + R_{2} = w$ (Balancing Forces in $Y - a x$ ) __(I)
Writing equation for torque about the centre of bar :
$R_{1} (0.5) + w (0) - R_{2} (0.5) = 0$
$R_{1} = R_{2}$ __(II)
From eq. (I) & (II)
$R_{1} = R_{2} = \frac{w}{2}$ Kgf.

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In Fig. 1.35, a uniform bar of length l m is supported at its end and loaded by a weight W kgf at its middle. In equilibrium, find the reactions R1 and R2 at the ends.

R1+R2=w (Balancing Forces in Y−ax) __(I)Writing equation for torque about the centre of bar :R1(0.5)+w(0)−R2(0.5)=0R1=R2 __(II)From eq. (I) & (II)R1=R2=w2Kgf.

In Fig. 1.35, a uniform bar of length l m is supported at its end and loaded by a weight W kgf at its middle. In equilibrium, find the reactions $R_{1}$ and $R_{2}$ at the ends.

$R_{1} + R_{2} = w$ (Balancing Forces in $Y - a x$ ) __(I)
Writing equation for torque about the centre of bar :
$R_{1} (0.5) + w (0) - R_{2} (0.5) = 0$
$R_{1} = R_{2}$ __(II)
From eq. (I) & (II)
$R_{1} = R_{2} = \frac{w}{2}$ Kgf.