Problem Overview
We decompose a 3D structural tension force vector F ⃗ = ⟨ 12 , 8 , 24 ⟩ kN \vec{F} = \langle 12, 8, 24 \rangle\text{ kN} F = ⟨ 12 , 8 , 24 ⟩ kN u ⃗ = ⟨ 2 , 1 , 2 ⟩ \vec{u} = \langle 2, 1, 2 \rangle u = ⟨ 2 , 1 , 2 ⟩
Step-by-Step Mathematical Solution
Step 1: Compute Vector Projection along the Strut (F ⃗ ∥ \vec{F}_{\parallel} F ∥
State the Vector Projection Formula :
proj u ⃗ F ⃗ = ( F ⃗ ⋅ u ⃗ ∥ u ⃗ ∥ 2 ) u ⃗ \text{proj}_{\vec{u}}\vec{F} = \left( \frac{\vec{F} \cdot \vec{u}}{\|\vec{u}\|^2} \right) \vec{u} proj u F = ( ∥ u ∥ 2 F ⋅ u ) u (This formula is defined in Vector Projections .)
Calculate the dot product:
F ⃗ ⋅ u ⃗ = ( 12 ) ( 2 ) + ( 8 ) ( 1 ) + ( 24 ) ( 2 ) = 24 + 8 + 48 = 80 kN \vec{F} \cdot \vec{u} = (12)(2) + (8)(1) + (24)(2) = 24 + 8 + 48 = 80\text{ kN} F ⋅ u = ( 12 ) ( 2 ) + ( 8 ) ( 1 ) + ( 24 ) ( 2 ) = 24 + 8 + 48 = 80 kN
Calculate the squared magnitude of u ⃗ \vec{u} u ∥ u ⃗ ∥ 2 = ( 2 ) 2 + ( 1 ) 2 + ( 2 ) 2 = 4 + 1 + 4 = 9 \|\vec{u}\|^2 = (2)^2 + (1)^2 + (2)^2 = 4 + 1 + 4 = 9 ∥ u ∥ 2 = ( 2 ) 2 + ( 1 ) 2 + ( 2 ) 2 = 4 + 1 + 4 = 9
Construct the parallel force vector:
F ⃗ ∥ = 80 9 ⟨ 2 , 1 , 2 ⟩ = ⟨ 160 9 , 80 9 , 160 9 ⟩ kN \vec{F}_{\parallel} = \frac{80}{9}\langle 2, 1, 2 \rangle = \left\langle \frac{160}{9}, \frac{80}{9}, \frac{160}{9} \right\rangle\text{ kN} F ∥ = 9 80 ⟨ 2 , 1 , 2 ⟩ = ⟨ 9 160 , 9 80 , 9 160 ⟩ kN
Step 2: Compute Orthogonal Shear Force Component (F ⃗ ⊥ \vec{F}_{\perp} F ⊥
State the Orthogonal Vector Component Formula :
F ⃗ ⊥ = orth u ⃗ F ⃗ = F ⃗ − proj u ⃗ F ⃗ \vec{F}_{\perp} = \text{orth}_{\vec{u}}\vec{F} = \vec{F} - \text{proj}_{\vec{u}}\vec{F} F ⊥ = orth u F = F − proj u F
F ⃗ ⊥ = ⟨ 12 , 8 , 24 ⟩ − ⟨ 160 9 , 80 9 , 160 9 ⟩ = ⟨ 108 − 160 9 , 72 − 80 9 , 216 − 160 9 ⟩ = ⟨ − 52 9 , − 8 9 , 56 9 ⟩ kN \vec{F}_{\perp} = \langle 12, 8, 24 \rangle - \left\langle \frac{160}{9}, \frac{80}{9}, \frac{160}{9} \right\rangle = \left\langle \frac{108 - 160}{9}, \frac{72 - 80}{9}, \frac{216 - 160}{9} \right\rangle = \left\langle -\frac{52}{9}, -\frac{8}{9}, \frac{56}{9} \right\rangle\text{ kN} F ⊥ = ⟨ 12 , 8 , 24 ⟩ − ⟨ 9 160 , 9 80 , 9 160 ⟩ = ⟨ 9 108 − 160 , 9 72 − 80 , 9 216 − 160 ⟩ = ⟨ − 9 52 , − 9 8 , 9 56 ⟩ kN Check orthogonality via dot product:
F ⃗ ⊥ ⋅ u ⃗ = ( − 52 9 ) ( 2 ) + ( − 8 9 ) ( 1 ) + ( 56 9 ) ( 2 ) = − 104 − 8 + 112 9 = 0 ✓ \vec{F}_{\perp} \cdot \vec{u} = \left(-\frac{52}{9}\right)(2) + \left(-\frac{8}{9}\right)(1) + \left(\frac{56}{9}\right)(2) = \frac{-104 - 8 + 112}{9} = 0\quad \checkmark F ⊥ ⋅ u = ( − 9 52 ) ( 2 ) + ( − 9 8 ) ( 1 ) + ( 9 56 ) ( 2 ) = 9 − 104 − 8 + 112 = 0 ✓ Step 3: Magnitudes of Axial and Shear Loads
Axial Compressive Load:
∥ F ⃗ ∥ ∥ = 80 9 ∥ ⟨ 2 , 1 , 2 ⟩ ∥ = 80 9 ( 3 ) = 80 3 ≈ 26.67 kN \|\vec{F}_{\parallel}\| = \frac{80}{9}\|\langle 2, 1, 2 \rangle\| = \frac{80}{9}(3) = \frac{80}{3} \approx 26.67\text{ kN} ∥ F ∥ ∥ = 9 80 ∥ ⟨ 2 , 1 , 2 ⟩ ∥ = 9 80 ( 3 ) = 3 80 ≈ 26.67 kN Lateral Shear Load:
∥ F ⃗ ⊥ ∥ = ∥ F ⃗ ∥ 2 − ∥ F ⃗ ∥ ∥ 2 \|\vec{F}_{\perp}\| = \sqrt{\|\vec{F}\|^2 - \|\vec{F}_{\parallel}\|^2} ∥ F ⊥ ∥ = ∥ F ∥ 2 − ∥ F ∥ ∥ 2 ∥ F ⃗ ∥ 2 = ( 12 ) 2 + ( 8 ) 2 + ( 24 ) 2 = 144 + 64 + 576 = 784 ⟹ ∥ F ⃗ ∥ = 28 kN \|\vec{F}\|^2 = (12)^2 + (8)^2 + (24)^2 = 144 + 64 + 576 = 784 \implies \|\vec{F}\| = 28\text{ kN} ∥ F ∥ 2 = ( 12 ) 2 + ( 8 ) 2 + ( 24 ) 2 = 144 + 64 + 576 = 784 ⟹ ∥ F ∥ = 28 kN ∥ F ⃗ ⊥ ∥ = 784 − 6400 9 = 7056 − 6400 9 = 656 9 = 4 41 3 ≈ 8.54 kN \|\vec{F}_{\perp}\| = \sqrt{784 - \frac{6400}{9}} = \sqrt{\frac{7056 - 6400}{9}} = \sqrt{\frac{656}{9}} = \frac{4\sqrt{41}}{3} \approx 8.54\text{ kN} ∥ F ⊥ ∥ = 784 − 9 6400 = 9 7056 − 6400 = 9 656 = 3 4 41 ≈ 8.54 kN
Step 4: Compute the Overturning Moment (Torque) about the Footing
State the Torque Vector Formula :
τ ⃗ = r ⃗ × F ⃗ \vec{\tau} = \vec{r} \times \vec{F} τ = r × F (This formula is defined in Torque and Rotational Mechanics .)
τ ⃗ = ∣ i ^ j ^ k ^ 6 4 12 12 8 24 ∣ \vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 4 & 12 \\ 12 & 8 & 24 \end{vmatrix} τ = i ^ 6 12 j ^ 4 8 k ^ 12 24 F ⃗ = 2 r ⃗ \vec{F} = 2\vec{r} F = 2 r r ⃗ \vec{r} r τ ⃗ = ⟨ ( 4 ) ( 24 ) − ( 12 ) ( 8 ) , ( 12 ) ( 12 ) − ( 6 ) ( 24 ) , ( 6 ) ( 8 ) − ( 4 ) ( 12 ) ⟩ = ⟨ 0 , 0 , 0 ⟩ kN ⋅ m \vec{\tau} = \langle (4)(24) - (12)(8), \; (12)(12) - (6)(24), \; (6)(8) - (4)(12) \rangle = \langle 0, 0, 0 \rangle\text{ kN}\cdot\text{m} τ = ⟨( 4 ) ( 24 ) − ( 12 ) ( 8 ) , ( 12 ) ( 12 ) − ( 6 ) ( 24 ) , ( 6 ) ( 8 ) − ( 4 ) ( 12 )⟩ = ⟨ 0 , 0 , 0 ⟩ kN ⋅ m Engineering Conclusion
Because the stay cable pulls collinear with position vector r ⃗ \vec{r} r 0 kN ⋅ m 0\text{ kN}\cdot\text{m} 0 kN ⋅ m 26.67 kN 26.67\text{ kN} 26.67 kN 8.54 kN 8.54\text{ kN} 8.54 kN