### Problem Statement**Objective:** Classify the truss and find the indicated bar force, as well as the reactions.### Diagram ExplanationThe diagram is a schematic of a truss structure comprising multiple members and joints. Here's a detailed explanation:- **Truss Members and Joints:** - The truss consists of joints labeled A, B, C, D, E, F, and G. - At joint A, a force of 20 kips (20k) is applied. - The truss spans horizontally from joint A to joint G, across six segments.- **Supports:** - There is a hinged support at joint A (represented by a triangle), allowing rotation but preventing translation. - A roller support is present at joint G, allowing horizontal movement but resisting vertical forces.- **Vertical Loads:** - Joint B experiences a vertical downward load of 5 kips (5k). - Joint C carries a vertical downward load of 10 kips (10k). - Joint D experiences a vertical downward load of 10 kips (10k). - Joint E carries a vertical downward load of 5 kips (5k). - Joint F holds a vertical downward load of 5 kips (5k).- **Dimensions:** - The total horizontal span covered by the joints from A to G is 150 feet, divided into six equal segments of 25 feet each. - The vertical height from the base line to the top joints is 30 feet.By analyzing the truss structure, support conditions, and applied loads, the task is to classify the type of truss and calculate the internal bar forces and external reactions.

Consider the following free body diagram: The above truss is a two dimensional truss. Check for static determinacy: D = m + r - 2j Here, m is the number of members, r is the number of reactions, and j is the number of joints. Substitute, m = 25, r = 3, and j = 14. D = 25 + 3 - (2x14) D = 25 + 3 - 28 D = 0 (statically determinate structure)Consider the equilibrium equations as follows: ∑Ma = 0(gy×150) - (5×125) -(5×100) -(10×75)-(10×50)-(5×25)-(20×30)=0 gy = 20.667 k∑Fx = 0-ax + 20 = 0 ax = -20 k∑Fy = 0ay + gy -5 -10 -10 -5 -5 = 0ay + gy -35 = 0Substitute, gy = 20.667 kay + 20.667 -35 = 0 ay = 14.333 kConsider the following section from the truss: Calculate the value of θ θ = tan-13050θ = tan-10.6θ = 30.964 ∑ Fy = 0ay -Fbd sinθ -5 = 0Substitute ay = 14.333 k and θ =30.96414.333 -Fbd sin(30.964) -5 = 0 Fbd = 18.141 k (Tension) Therefore, the value of force in member BD = 18.141 k