Let's break down **Problem 1** as requested, in two steps with explanations after each step, and then provide the final answer for each part. --- ## **Problem 1** ### **Step 1: Identify Zero-Force Members** #### **Explanation:** - **Zero-force members** are found using two basic rules: 1. If two non-collinear members meet at a joint unstressed by external force or support, both are zero-force members. 2. If three members form a truss joint and two are collinear, the third member is a zero-force member if no external/load/support acts at the joint. - **Apply Rule 1 & 2 to the truss:** - **Joint F:** Members \( EF, FK, FG \). No load/support at F, and EF and FG are collinear. By Rule 2, **FK is a zero-force member.** - **Joint H:** Members \( GH, HM, HI \). No load/support at H, and HM and GH are collinear. By Rule 2, **HI is a zero-force member.** - **Joint D:** Members \( CD, DE, DF \). No load/support at D, and DE and CD are not collinear, so **no zero-force member here.** - **Joint C:** Similar analysis, but since it connects to support, no zero-force member. - **Joints at supports (A, B, I):** Don't apply zero-force rules due to supports. **Zero-force members:** - \( FK \) - \( HI \) --- ### **Step 2: Find the Reactions** #### **Explanation:** - The truss has pin support at \( A \) and roller support at \( I \). - External loads: \( 1 \) kip at \( K \), \( 2 \) kip at \( L \), \( 1 \) kip at \( M \). **Let:** - \( A_y \) and \( A_x \) be reactions at \( A \) (vertical and horizontal), - \( I_y \) be vertical reaction at \( I \). **Sum of moments about \( A \):** \[ \sum M_A = \] - Distance from \( A \) to \( K \): \( 8 + 8 + 12 = 28 \) ft - Distance from \( A \) to \( L \): \( 8 + 8 + 12 + 12 = 40 \) ft - Distance from \( A \) to \( M \): \( 8 + 8 + 12 + 12 + 12 = 52 \) ft - Distance from \( A \) to \( I \): \( 8 + 8 + 12 + 12 + 12 = 52 \) ft \[ 1 \text{ kip} \times 28 \text{ ft} + 2 \text{ kip} \times 40 \text{ ft} + 1 \text{ kip} \times 52 \text{ ft} - I_y \times 52 \text{ ft} = \] \[ (28) + (80) + (52) - 52I_y = \] \[ 160 = 52I_y \implies I_y = \frac{160}{52} \approx 3.08 \text{ kip} \] **Sum of vertical forces:** \[ A_y + I_y = 1 + 2 + 1 = 4 \text{ kip} \] \[ A_y + 3.08 = 4 \implies A_y = .92 \text{ kip} \] **Sum of horizontal forces:** - No horizontal loads, so \( A_x = \). --- ### **Final Answers** #### **a) Zero-force members:** - **FK** and **HI**. #### **b) Reactions:** - \( A_x = \) kip - \( A_y = .92 \) kip (upward) - \( I_y = 3.08 \) kip (upward) --- If you'd like to continue to parts (c) and (d), let me know!

## **Problem 1** ### **Step 1: Identify Zero-Force Members** #### **Explanation:** - **Zero-force members** are determined using two key rules: 1. If two non-collinear members meet at a joint with no external load, both members are zero-force members. 2. If three members form a joint where two are collinear, the third member is a zero-force member if no external load is applied at that joint. - **Analysis:** - **Joint F:** Members \( EF, FK, FG \) — No external load, and \( EF \) and \( FG \) are collinear. Therefore, **FK is a zero-force member.** - **Joint H:** Members \( GH, HM, HI \) — No external load, and \( GH \) and \( HM \) are collinear. Thus, **HI is a zero-force member.** - **Joint D:** Members \( CD, DE, DF \) — No zero-force members found as none of the conditions apply. - **Joint C and others:** Similar reasoning shows no additional zero-force members. **Zero-force members identified:** - **FK** - **HI** --- ### **Step 2: Find the Reactions** #### **Explanation:** - The truss is supported at \( A \) (pin support) and \( I \) (roller support). - External loads: \( 1 \) kip at \( K \), \( 2 \) kip at \( L \), and \( 1 \) kip at \( M \). **Let:** - \( A_y \) = vertical reaction at \( A \) - \( A_x \) = horizontal reaction at \( A \) (assumed zero) - \( I_y \) = vertical reaction at \( I \) **Sum of moments about point \( A \):** \[ \sum M_A = 0 \] \[ 1 \text{ kip} \times 28 \text{ ft} + 2 \text{ kip} \times 40 \text{ ft} + 1 \text{ kip} \times 52 \text{ ft} - I_y \times 52 \text{ ft} = 0 \] Calculating distances: - \( A \) to \( K \): \( 28 \) ft - \( A \) to \( L \): \( 40 \) ft - \( A \) to \( M \): \( 52 \) ft The equation becomes: \[ 28 + 80 + 52 - 52I_y = 0 \] \[ 160 = 52I_y \implies I_y = \frac{160}{52} \approx 3.08 \text{ kip} \] **Sum of vertical forces:** \[ A_y + I_y = 4 \text{ kip} \] Substituting \( I_y \): \[ A_y + 3.08 = 4 \implies A_y = 4 - 3.08 \implies A_y \approx 0.92 \text{ kip} \] **Horizontal forces:** - Since no horizontal loads are applied, \( A_x = 0 \). --- ### **Final Answers for Problem 1** #### **a) Zero-force members:** - **FK** - **HI** #### **b) Reactions:** - \( A_x = 0 \) kip - \( A_y \approx 0.92 \) kip (upward) - \( I_y \approx 3.08 \) kip (upward) --- ## **Problem 2** ### **Step 1: Compute Forces at Joints C, E, and G** #### **Explanation:** - The wind pressure creates a uniform distributed load \( w = 1 \) kip/ft. This load is applied to the truss panel and transferred to the joints. **Total Load on Members:** - Given the length of the panel, if it spans 6 ft (assumed based on the image), the total load is: \[ w_{\text{total}} = w \times \text{length} = 1 \text{ kip/ft} \times 6 \text{ ft} = 6 \text{ kips} \] This load is distributed among joints \( C, E, G \). **Assuming symmetry:** - Each joint (C, E, G) will carry \( 2 \) kips if equally distributed (3 joints). **Forces at Joints:** - Joint \( C \): \( 2 \) kips - Joint \( E \): \( 2 \) kips - Joint \( G \): \( 2 \) kips --- ### **Step 2: Determine Internal Forces in Members ED and CD** #### **Explanation:** - Use the method of joints to find the internal forces in the members. **Joint D Analysis:** 1. **For Member ED:** - Since joint D is under tension due to the upward load from joints C and E, member \( ED \) is in **tension**. 2. **For Member CD:** - Member \( CD \) is also under tension as it connects to joint C where the load is acting, so \( CD \) is in **tension** as well. **Final Forces:** - **Internal force in member ED:** Tension - **Internal force in member CD:** Tension --- ### **Final Answers for Problem 2** #### **a) Forces at Joints C, E, and G:** - Each joint carries approximately \( 2 \) kips. #### **b) Internal force in member ED:** - **Tension.** #### **c) Internal force in member CD:** - **Tension.** If you need further analysis or details, please let me know!