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4–109. Replace the loading by an equivalent resultant force and couple moment at point O.
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4–109. Replace the loading by an equivalent resultant force and couple moment at point O.
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Step 1: Find the Line Vectors \( \mathbf{OA} \) and \( \mathbf{OB} \)
We are given the points \( O(0, 0, 0) \), \( A(0.8, 0, -1.2) \), and \( B(0, -0.5, 0) \). The position vectors from point \( O \) to points \( A \) and \( B \) are calculated as follows:
The vector \( \mathbf{OA} \) is:
\[
\mathbf{OA} = \{ (0.8 – 0) \hat{i} + (0 – 0) \hat{j} + (-1.2 – 0) \hat{k} \} = \{ 0.8 \hat{i} + 0 \hat{j} – 1.2 \hat{k} \} \, \text{m}
\]
Similarly, the vector \( \mathbf{OB} \) is:
\[
\mathbf{OB} = \{ (0 – 0) \hat{i} + (-0.5 – 0) \hat{j} + (0 – 0) \hat{k} \} = \{ 0 \hat{i} – 0.5 \hat{j} + 0 \hat{k} \} \, \text{m}
\]
Step 2: Calculate the Resultant Force
The forces \( \mathbf{F_1} \) and \( \mathbf{F_2} \) are given as:
\[
\mathbf{F_1} = \{ 8 \hat{i} – 2 \hat{k} \} \, \text{kN}
\]
\[
\mathbf{F_2} = \{ -2 \hat{i} + 5 \hat{j} – 3 \hat{k} \} \, \text{kN}
\]
The resultant force \( \mathbf{F_R} \) is the sum of the two forces \( \mathbf{F_1} \) and \( \mathbf{F_2} \):
\[
\mathbf{F_R} = \mathbf{F_1} + \mathbf{F_2}
\]
Substituting the values of \( \mathbf{F_1} \) and \( \mathbf{F_2} \):
\[
\mathbf{F_R} = \{ 8 \hat{i} – 2 \hat{k} \} + \{ -2 \hat{i} + 5 \hat{j} – 3 \hat{k} \}
\]
\[
\mathbf{F_R} = \{ 6 \hat{i} + 5 \hat{j} – 5 \hat{k} \} \, \text{kN}
\]
Step 3: Calculate the Moment Due to Force \( \mathbf{F_1} \) About Point \( O \)
The moment due to force \( \mathbf{F_1} \) about point \( O \) is calculated as the cross product of the position vector \( \mathbf{OA} \) and the force vector \( \mathbf{F_1} \):
\[
\mathbf{M_1} = \mathbf{OA} \times \mathbf{F_1}
\]
Substituting the values:
\[
\mathbf{M_1} = \left| \begin{matrix}
\hat{i} & \hat{j} & \hat{k} \\
0.8 & 0 & -1.2 \\
8 & 0 & -2
\end{matrix} \right|
\]
\[
\mathbf{M_1} = \{ 0 \hat{i} – 8 \hat{j} + 9.6 \hat{k} \} \, \text{kN} \cdot \text{m}
\]
Step 4: Calculate the Moment Due to Force \( \mathbf{F_2} \) About Point \( O \)
The moment due to force \( \mathbf{F_2} \) about point \( O \) is calculated as the cross product of the position vector \( \mathbf{OB} \) and the force vector \( \mathbf{F_2} \):
\[
\mathbf{M_2} = \mathbf{OB} \times \mathbf{F_2}
\]
Substituting the values:
\[
\mathbf{M_2} = \left| \begin{matrix}
\hat{i} & \hat{j} & \hat{k} \\
0 & -0.5 & 0 \\
-2 & 5 & -3
\end{matrix} \right|
\]
\[
\mathbf{M_2} = \{ 2.5 \hat{i} + 1 \hat{j} + 0 \hat{k} \} \, \text{kN} \cdot \text{m}
\]
Step 5: Calculate the Resultant Moment
The resultant moment about point \( O \) is the sum of the moments \( \mathbf{M_1} \) and \( \mathbf{M_2} \):
\[
\mathbf{M_O} = \mathbf{M_1} + \mathbf{M_2}
\]
Substituting the values of \( \mathbf{M_1} \) and \( \mathbf{M_2} \):
\[
\mathbf{M_O} = \{ 0 \hat{i} – 8 \hat{j} + 9.6 \hat{k} \} + \{ 2.5 \hat{i} + 1 \hat{j} + 0 \hat{k} \}
\]
\[
\mathbf{M_O} = \{ 2.5 \hat{i} – 7 \hat{j} + 9.6 \hat{k} \} \, \text{kN} \cdot \text{m}
\]
Conclusion:
The equivalent system consists of a resultant force \( \mathbf{F_R} = \{ 6 \hat{i} + 5 \hat{j} – 5 \hat{k} \} \, \text{kN} \) and a resultant moment \( \mathbf{M_O} = \{ 2.5 \hat{i} – 7 \hat{j} + 9.6 \hat{k} \} \, \text{kN} \cdot \text{m} \).
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