please specify if the results are compression or tension The image depicts a truss structure used in engineering mechanics to illustrate force distribution in a framework. The truss consists of several interconnected members forming triangles. Key dimensions and angles are provided to assist in analysis:- **Members and Joints**: The structure has joints labeled A, B, C, D, and E. Each member is formed by a straight connection between two joints.- **Triangles**: Two triangular sections are prominent, one between joints A, B, and E, and the other between joints E, C, and D.- **Dimensions**: Horizontal members AB, BC, and CD measure 4 meters each.- **Angles**: The angle between member AE and the horizontal member AB is 60°. Similarly, the angle between member ED and the horizontal member CD is 60°.- **Support and Load**: Joint A is supported at the base, and a downward load labeled P is applied at joint D.- **Horizontal Lengths**: The total horizontal span from A to D is 12 meters, divided into three segments of 4 meters each.This diagram aids in understanding the distribution of forces within a trussed structure and is essential for analyzing structural integrity and stability. ### Problem 4#### 4.aDetermine the force in each member of the truss. State whether the members are in tension or compression. Set \( P = 8 \) kN.#### 4.bIf the maximum force that any member can support is 8 kN in tension and 6 kN in compression, determine the maximum force \( P \) that can be supported at joint \( D \).**Hint:** Forces in members are directly proportional with force \( P \).

Answered: Image /qna-images/answer/e6708d99-d3b0-490f-a48d-478f924d1e53.jpg∑MA = 08×P - 4×RE = 064 - 4RE = 0RE = 16 kN∑Fy = 0RA + RE = 8RA = 8 -RERA = 8 -16 = -8 kN ∑Fx = 0HA = 0Joint A and Joint B ∑Fy = 0RA + FABsin60 = 0FAB = -RAsin60FAB = --8sin60FAB = 9.24 kN∑Fx = 0FABcos60 +FAE = 0FAE = - FABcos60FAE = -9.24cos60FAE = -4.62 kNJoint B∑Fy = 0FBEcos30 +FABcos30 = 0FBE = -FABFBE = -9.24 kN∑Fx = 0FCB = -FBEcos60FCB = --9.24cos60FCB = 4.62 kN Joint C and Joint D Joint D∑Fy = 0FCDsin60=8FCD = 8sin60FCD = 9.24 kN∑Fx = 0FCDcos60 +FED = 0FED = -9.24×cos60FED = -4.62 kN Joint C∑Fy = 0FCDcos30+FCEcos30 = 0FCD = -FCEFCE =-9.24 kN∑Fx = 0FCB = FCDcos60FCB = 9.24 cos60FCB = 4.62 kNMember Magnitude Nature AB 9.24 T AE 4.62 C BE 9.24 C BC 4.62 T ED 4.62 C CE 9.24 C CD 9.24 TMaximum compression force = 6 kN Maximum tensile force = 8 kN Force at P in original truss = 8kN From the table, The maximum tensile force in the original truss = 9.24 kN As force in members are directly proportional to PP8=89.62P =6.65 kN