Please solve this problem in detail so a beginner can understand your thought process. Go over every step. Subject is Structural Analysis **Influence Line Analysis for Truss Members****Question 6:**Draw the influence line for the force in:(a) Member JI(b) Member IE(c) Member EF**Solution:**The problem involves evaluating a truss structure, specifically identifying the influence lines for forces in certain members as a load traverses the truss.**Truss Diagram:**The truss consists of members labeled A through G, with verticals increasing by 6 ft per section. The height of the truss is 8 ft.**(a) Influence Line for Member JI:**- Graph is plotted on a coordinate system where the x-axis represents the position of the moving load in feet, and the y-axis represents the influence line values.- Points are noted at: - (0, 0) - (6, -0.25) - (12, -0.5) - (18, -0.75) - (24, -1) - (30, -0.5) - (36, 0)- The line descends linearly from the origin to a minimum value at x = 24 ft, then returns to zero at x = 36 ft.**(b) Influence Line for Member IE:**- Graph is plotted with similar axes where x represents feet and y the influence value.- Points are noted at: - (0, 0) - (6, 0.167) - (12, 0.333) - (18, 0.5) - (24, 0.667) - (28.8, 0) - (30, -0.167) - (36, 0)- The graph peaks at x = 24 ft before dropping to a zero and then a slight negative value at x = 30 ft.**(c) Influence Line for Member EF:**- Graph depicts a symmetric influence line.- Points are noted at: - (0, 0) - (6, 0.25) - (12, 0.5) - (18, 0.75) - (24, 1.0) - (30, 0.5) - (36, 0)- The line increases to a maximum at x = 24 ft

Given Data: Total span=36 ftH=8 ftPannel span =6 ft eachTo determine Influence Line for Member JI, IE, EF To find:- Influence diagram of member JI, IE and EFSupport reactions:- Lets assume that the moving unit load of 1 kip is at a distance of x from support A Taking sum of moments at G equal to zero ∑MG=0RA×36-1×36-x=0RA=36-x36RA=1-x36 Taking sum of vertical forces equal to zero ∑Fy=0RA+RG-1=01-x36+RG-1=0RG=x36(a) Influence line for member JI:- Take a section aa through members JI, IE and EF When the unit load is at left side of the section aa 0⩽x⩽24 ft, taking right part of the section Taking sum of moments at E equal to zero ∑ME=0FJI×8+RG×12=0FJI=-RG×128FJI=-x36×128FJI=-x24 At x=0FJI=-024=0At x=6 ft,FJI=-624=-0.25At x=12 ft,FJI=-1224=-0.5At x=18 ft,FJI=-1824=-0.75At x=24 ft,FJI=-2424=-1 When the unit load is at right side of the section aa 30⩽x⩽36 ft, taking left part of the section Taking sum of moments at E equal to zero ∑ME=0RA×24+FJI×8=0FJI=-RA×248FJI=-1-x36×248FJI=-1-x36×3FJI=-3+x12 At x=30 ft,FJI=-3+3012=-0.5At x=36 ft,FJI=-3+3612=0 Influence line for member JI is given as(b) Influence line for member IE:- When the unit load is at left side of the section aa 0⩽x⩽24 ft, taking right part of the section Taking sum of vertical forces equal to zero ∑Fy=0-FIE+RG=0FIE=x36 At x=0,FIE=036=0At x=6 ft,FIE=636=0.167At x=12 ftFIE=1236=0.333At x=18 ft,FIE=1836=0.5At x=24 ft,FIE=2436=0.667 When the unit load is at right side of the section aa 30⩽x⩽36 ft, taking left part of the section Taking sum of vertical forces equal to zero ∑Fy=0FIE+RA=0FIE=-RAFIE=-(1-x36) At x=30 ft,FIE=-1-3036=-0.167At x=36ft,FIE=-1-3636=0 Influence line of member IE is given as(c) Influence line for member EF:- Take a section aa through members JI, IE and EF When the unit load is at left side of the section aa 0⩽x⩽24 ft, taking right part of the section Taking sum of moments at I equal to zero ∑MI=0RG×12-FEF×8=0FEF=RG×128FEF=x36×128FEF=x24 At x=0,FEF=024=0At x=6 ft,FEF=624=0.25At x=12 ftFEF=1224=0.5At x=18 ftFEF=1824=0.75At x=245 ftFEF=2424=1 When the unit load is at right side of the section aa 30⩽x⩽36 ft, taking left part of the section Taking sum of moments at I equal to zero ∑MI=0RA×24-FEF×8=0FEF=RA×248FEF=1-x36×248FEF=3-x12 At x=30 ft,FEF=3-3012=0.5At x=36 ftFEF=3-3612=0 Influence line of member EF is given as