# Step-by-Step Solution: Average Normal Stress in Truss Bars ## Problem Overview - **Given:** - Each truss bar has a **diameter \( d = 20\, \text{mm} \)** - **Load \( P = 50\, \text{kN} \)** - Truss geometry: - \( AB = 2.\, \text{m} \) - \( BC = 1.5\, \text{m} \) - Assume simple pin-jointed triangle with joints at \( A \) (left), \( B \) (bottom right), and \( C \) (top). - **Find:** - The **average normal stress** in bars \( BC \), \( AC \), and \( AB \). --- ## 1. **Draw the Truss and Assign Joints** ``` C |\ | \ |__\ A B ``` - \( AB = 2.\, \text{m} \) - \( BC = 1.5\, \text{m} \) - \( AC = \sqrt{(2^2 + 1.5^2)} = 2.5\, \text{m} \) (by Pythagoras) --- ## 2. **Find Support Reactions** Assume: - Pin at \( A \), roller at \( B \) - Load \( P = 50\, \text{kN} \) applied vertically downward at joint \( C \) **Sum of vertical forces:** \[ \sum F_y = \implies A_y + B_y = P \] **Moment about A:** \[ \text{Distance from A to C horizontally} = 2\,\text{m} \] \[ \text{Distance from B to C horizontally} = \] \[ \sum M_A = : \quad B_y \times 2. - 50\,\text{kN} \times 2. = \] \[ B_y = 50\,\text{kN} \] \[ A_y + 50 = 50 \implies A_y = \] --- ## 3. **Joint C: Find Bar Forces** At joint \( C \), two unknown bar forces: - \( F_{AC} \) (along \( AC \)) - \( F_{BC} \) (along \( BC \)) - Load \( P \) downward ### Step 1: Geometry for Direction Cosines - \( AC \) runs from \( A(,) \) to \( C(2, 1.5) \) - \( BC \) runs from \( B(2,) \) to \( C(2, 1.5) \) - \( AB \) runs from \( A(,) \) to \( B(2,) \) So: - \( AC \): horizontal \( = +2 \), vertical \( = +1.5 \) - \( BC \): horizontal \( = \), vertical \( = +1.5 \) ### Step 2: Free-Body Diagram at C Let: - \( F_{AC} \) acts **down and left** (assume tension) - \( F_{BC} \) acts **down** (assume tension) - Load \( P = 50\,\text{kN} \) acts **down** #### Resolve Forces at C: \[ \sum F_x = : \quad F_{AC} \frac{2}{2.5} = \implies F_{AC} \text{ has only x-component} \] But in vertical: \[ \sum F_y = : \quad F_{AC} \frac{1.5}{2.5} + F_{BC} - 50 = \] But better to write both equations: ##### \( F_{AC} \) direction cosines: - \( l_{AC,x} = \frac{2}{2.5} = .8 \) - \( l_{AC,y} = \frac{1.5}{2.5} = .6 \) ##### \( F_{BC} \): - Only vertical (since B and C have same x), so: - \( l_{BC,x} = \) - \( l_{BC,y} = 1 \) #### Horizontal equilibrium at C: \[ \sum F_x = : \quad F_{AC} \cdot .8 = \implies F_{AC} = \] But this can't be; let's double-check. Actually, at C, the only horizontal force is from AC. But let's approach via method of joints: --- ## 4. **Method of Joints: Start at Joint C** **At C:** - Unknowns: \( F_{AC} \) (along AC), \( F_{BC} \) (along BC) - Applied load: 50 kN down Write equilibrium equations: ### Vertical: \[ \sum F_y = : \quad F_{AC} \cdot \sin \theta + F_{BC} - 50 = \] ### Horizontal: \[ \sum F_x = : \quad F_{AC} \cdot \cos \theta = \implies F_{AC} = \] So, all vertical component is taken by BC. \[ F_{AC} = \] \[ F_{BC} = 50\,\text{kN} \] --- ## 5. **Joint B** At joint B: - \( F_{AB} \) (along AB, unknown) - \( F_{BC} \), already found (upward in BC) - Reaction \( B_y = 50\,\text{kN} \) upward ### Vertical equilibrium: \[ F_{BC} + B_y = \] But both are upwards, so for equilibrium, \( F_{BC} \) must act down at B, which matches the member being in tension. ### Horizontal equilibrium: No horizontal forces at B, so \[ F_{AB} = \] --- ## 6. **Summary of Bar Forces** - \( F_{AC} = \) - \( F_{BC} = 50\,\text{kN} \) (tension) - \( F_{AB} = \) --- ## 7. **Calculate Area of Each Bar** \[ A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (.02)^2 = 3.142 \times 10^{-4}\,\text{m}^2 \] --- ## 8. **Average Normal Stress Formula** \[ \sigma = \frac{F}{A} \] --- ## 9. **Final Answers** ### **Part A: Bar BC** \[ \sigma_{BC} = \frac{50,000\,\text{N}}{3.142 \times 10^{-4}\,\text{m}^2} = 159,155\,\text{Pa} = 159\,\text{MPa} \] **Sign:** Positive, since it's in tension. --- ### **Part B: Bar AC** \[ F_{AC} = \implies \sigma_{AC} = \] --- ### **Part C: Bar AB** \[ F_{AB} = \implies \sigma_{AB} = \] --- ## **Summary Table** | Bar | Force (kN) | Stress (MPa) | Sign | |-------|------------|--------------|-----------| | BC | 50 | +159 | Tension | | AC | | | -- | | AB | | | -- | --- ## **Final Answers (to 3 significant figures)** - **Bar BC:** \[ \boxed{159\,\text{MPa}} \] (Tension; enter as +159 MPa) - **Bar AC:** \[ \boxed{\,\text{MPa}} \] - **Bar AB:** \[ \boxed{\,\text{MPa}} \] --- ## **Truss Diagram**  *Alt text: Truss with nodes at A(,), B(2,), C(2,1.5), P=50kN at C, bar BC in tension.* --- ### **Key Points** - Only bar **BC** carries the load; other bars have zero force. - **Average normal stress** is only present in bar BC, with a value of **159 MPa (tension)**.