determine the reactions at the supports for the beam shown in fig. 3.16(a).

Explanation:

Step1: Resolve the 54 - kN force

The vertical component of the 54 - kN force is $F_{54y}=54\sin60^{\circ}=54\times\frac{\sqrt{3}}{2}\text{ kN}\approx46.77\text{ kN}$ and the horizontal component is $F_{54x}=54\cos60^{\circ}=27\text{ kN}$.

Step2: Sum of horizontal forces

Let the horizontal reaction at the left - hand support be $H$. Since there are no other horizontal forces, $\sum F_x = 0$, so $H - 27=0$, then $H = 27\text{ kN}$.

Step3: Sum of moments about the left - hand support

Let the vertical reaction at the right - hand support be $R_2$ and the vertical reaction at the left - hand support be $R_1$. $\sum M_1=0$. The moment of the 54 - kN vertical component about the left - hand support is $46.77\times3\text{ kN}\cdot\text{m}$, the moment of the 27 - kN force about the left - hand support is $27\times(3 + 3+1.5)\text{ kN}\cdot\text{m}$, and the moment of $R_2$ about the left - hand support is $R_2\times(3 + 3)\text{ kN}\cdot\text{m}$. $46.77\times3+27\times(3 + 3+1.5)-R_2\times(3 + 3)=0$. $140.31+27\times7.5-6R_2=0$. $140.31 + 202.5-6R_2=0$. $342.81-6R_2=0$. $R_2=\frac{342.81}{6}=57.14\text{ kN}$.

Step4: Sum of vertical forces

$\sum F_y = 0$, so $R_1+R_2-46.77 - 27=0$. $R_1+57.14-46.77 - 27=0$. $R_1=46.77 + 27-57.14$. $R_1=16.63\text{ kN}$.

Answer:

Horizontal reaction at left - hand support: $27\text{ kN}$ Vertical reaction at left - hand support: $16.63\text{ kN}$ Vertical reaction at right - hand support: $57.14\text{ kN}$