Question

determine the components of the support reactions at the fixed support a on the cantilevered beam. (figure 1)

part b

determine the moment of reaction at point a.

express your answer to three significant figures and include the appropriate units.

Explanation:

Step1: Calculate moment due to 6 - kN force

The moment due to the 6 - kN force about point A is \(M_1=6\times(1.5 + 1.5)\) (using \(M = F\times d\), where \(d\) is the perpendicular distance from the force to the point).

$$M_1=6\times3=18\space kN\cdot m$$

Step2: Calculate moment due to 4 - kN force

The perpendicular distance from the 4 - kN force to point A:
The horizontal component of the distance for the 4 - kN force: \(d_{x}=1.5\sin30^{\circ}+1.5 + 1.5\) and the vertical component \(d_{y}=1.5\cos30^{\circ}\). Using the formula \(M = F\times d\) (where \(d\) is the perpendicular distance). The moment due to the 4 - kN force about point A is \(M_2 = 4\times(1.5\sin30^{\circ}+1.5 + 1.5)\cos30^{\circ}+4\times1.5\cos30^{\circ}\sin30^{\circ}\)

Another way: Using the principle of moments \(M=\sum F\times d\). The total moment about point A is \(M_A\).
The moment due to 6 - kN force: \(M_{6}=6\times(1.5 + 1.5)=18\space kN\cdot m\)
The moment due to 4 - kN force: The horizontal distance from 4 - kN force to A is \(x=(1.5 + 1.5+1.5\sin30^{\circ})\) and vertical distance \(y = 1.5\cos30^{\circ}\). The moment of 4 - kN force about A is \(M_{4}=4\times(1.5 + 1.5+1.5\sin30^{\circ})\cos30^{\circ}+4\times1.5\cos30^{\circ}\sin30^{\circ}\)

Simplify \(M_{4}\):

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The total moment \(M_A=M_{6}+M_{4}\)

$$
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Answer:

\(M_A = 33.6\space kN\cdot m\)