Question Problem 4 Truss Analysis - Method of Sections s (25 pts.) Using the truss from problem 1, determine the reactions and member forces for this new loading scenario. The new forces are: Joints B, C and D Fy = 15kN; Joint G Fy=40kN; Joint E Fy- 20kn. Use the sections method. Ignore the self-weight at this time. fm NONEW WINE HD If fos run

Sul { }^{\wedge}By Method of Sections:-(1) 1 S K N(1) ~ U K N,(3) ~ I S K NLet A x, A_{y}, ix be cuppontleeactions at A and I respectivey.\begin{array}{r}\rightarrow \quad \Sigma M_{A}=0 \\-(20 \times 12)-(15 \times 9)-(15 \times 6)-(15 \times 1)-(40 \times 6) \\\quad+6 I_{-x}=0\end{array}(1)(2)\begin{aligned}\theta & =\tan ^{-1}\left(\frac{G}{12}\right) \\& =26.56 \mathrm{~s}^{\circ} \\\frac{A L}{A E} & =\frac{\Delta H}{\Delta E}=\frac{C G}{C E}=\frac{\Delta F}{D E} \\-B K & =4.5 \mathrm{~m} \\C G & =7 \mathrm{~m}\end{aligned}\begin{array}{l}\sum F_{x}=0 \\A_{x}+I_{x}=0 \Rightarrow A x+125=0 \Rightarrow A x=-125 \mathrm{kN} \\\sum F_{y}=0 \\A y=-15-15-15-20-40=0 \Rightarrow A y=105 \mathrm{kN} \\\alpha=\tan ^{-1}\left(\frac{B H}{A B}\right)=\tan ^{-1}\left(\frac{4.5}{3}\right) \\=56.31\end{array}Using method of sectionTake Lis of (1)-(1) section\begin{array}{l}\sum M_{R}=0 \\6 I_{x}+\left(F_{I_{n}} \cos 0\right) 6=0 \\(6 \times 12 s)+\left(F_{L_{n}} \cos 26 . G_{61}\right) 6=0 \\R_{C n}=-139.7 \sin \text { (c) } \\\sum F_{y}=0\end{array}\varepsilon f_{y}=0A y-F_{A M} \sin \alpha+F_{I n} \sin \theta=0105-F_{A H} \sin 56.31-(139.75 \sin 26.565)=0\begin{array}{l}F_{A H}=51.08 \mathrm{kN}(t) \\\beta=45^{\circ}\end{array}Take sution (2) - (2)\begin{aligned}& +\quad \sum M_{B}=0 \\& \left(F_{H G}\right) B H+I_{x} \times 6-(A y \times 3)=0 \\\Rightarrow & F_{H G} 4.5+(115 \times 6)-(105 \times 3)=0 \\& F_{H} h=-96.67 \mathrm{KN} \text { (c) (c) }\end{aligned}\begin{array}{l}\sum f_{x}=0 \\A_{x}+C_{x}+f_{B C}+f_{B C} \cos B+f_{x a} \cos \theta=0 \\\Rightarrow f_{B C}+66.14 \cos 45-96.67 \cos 26.565=0 \\F_{B C}=39.64 \mathrm{kN}(T) \\\end{array}Take Rns of suction (3) -(1)\begin{aligned}y=\tan ^{-1}\left(\frac{\partial f}{3}\right) & =\tan ^{-1}\left(\frac{1.5}{3}\right) \\& =26.565^{\circ}\end{aligned}\begin{array}{l} \sum F_{x}=0 \\A_{x}+F_{A B}+F_{A n} \cos \alpha+F_{I_{n}} \cos \theta \\+I_{x}=0\end{array}F_{A B}=96.706 \mathrm{kN}(\tau)\begin{array}{l}\frac{E F_{y}=0}{1 A_{y}-15-F D G \sin B+F_{H} \sin \theta 00} \\\qquad F_{B G}=66.14 \mathrm{kN}(T)\end{array}G\begin{aligned}\varepsilon M_{F} & =0 \\& -(20 \times 3)+\left(F_{C D} \times D F\right)=0 \\& =(-60)+\left(F_{C D} \times 41.5\right)=0 \Rightarrow F_{C D}=40 \mathrm{kN}(T) \\\varepsilon F_{X} & =0 \\& -F_{C D}-F_{C F} \cos Y-F_{G C} \cos \theta=0 \\& \left.\left.-40-F_{C F} \cos 26.56\right)-F_{G F} \cos 16.56\right)=0 \ldots\end{aligned}\begin{array}{l}\sum F y=0 \\=-15-20+F(F \sin y-F G F \sin \theta=0 \\\left.-35+F_{f} \sin 2(.56)-F_{G F} \sin 26.56\right)=0 \quad \cdots \text { (2) }\end{array}From (1) and (2)F_{F}=16.77 \mathrm{kN}(\tau), \quad F_{G F}=-61.5 \mathrm{kN} \mathrm{(C)}Now at Joint 0\begin{array}{l} \sum F_{x}=0 \\F_{D O}=F_{D E}=40 \mathrm{kN}(T) \\\sum F_{y=0} \\-1 S-F_{D F}=0 \Rightarrow F_{D F}=-1 \delta \mathrm{kN}(C)\end{array}Af Joint E,\begin{array}{l}\sum F_{y}=0 \\-20-F_{E F} \sin \theta=0 \\F_{E F}=\frac{-20}{\sin 26.56 \mathrm{~J}} \\F_{E F}=44.72 \mathrm{kN} \quad \text { (C) }\end{array}At Joint D,\begin{array}{l}\sum F_{y}=0 \\-15-F_{B n}-F_{B C} \sin B=0 \\\left.-15-F_{B n}-66.14 \sin 4\right)=0 \\\quad F_{B n}=-61.768 \mathrm{kN}\end{array}At Soint I\begin{array}{l}\angle F_{y}=0 \\F_{A_{I}}+F_{I n} \sin \theta=0 \\F_{A I}=-(139.75 \sin 26.56 \mathrm{~S})=0 \\F_{A_{I}}=62.5 \mathrm{KN} \quad(T)\end{array}At Joint C\begin{array}{l}\sum F_{y}=0 \\-F c_{r}-15 \cdot F_{C F} \sin y=0 \\-F c G-15-(16.72 \sin 26.565)=0 \\F\left(G_{r}=-22.51 \times N\right. \text { (c) }\end{array} ...