Question

Method of Sections Reference the truss shown below. Solve for the forces in the specified members using the Method of Sections. Clearly indicate whether the members are in compression (C) or tension (T). P-10 kN, P, = 6 kN and P, -8 kN. Note: You will not get credit for your work if you do not use the Method of Sections. 3 m Question 21 (6 points) Solve for the reactions at the supports Indicate both the magnitude and direction of these forces Question 22 14 points) Draw the section needed to solve for the forces in members DC. BE and ER Question 23 3 points) What is the force in member BC? Question 24 points What is the force in BE? Question 25 (3 points) What is the force in member EF?

Answer 1

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According to Given Question P₂=8KNF 021 For Balance of Vertical Reactions RFC D BX45 2m 3m RA+Ro = P + P2 RA 3m P = LOKN P₂ = 6 KN IRO RA+ Ro = 10+6 = 16 KN Now SMA =O (Moment about A) _=) Rox9-6x6-10x3 -8.x3 = 0 - Rox9=90 R = 10 KN RA = 16-Ro = 16-10 = 6 KN 22 These Sections are? P E t3 454 fr sm fat . B Fito 1823 Force r = force in BC = FBC from Section @ Take Moment about E, EME=0 SME = -Fc x S + Roxy=0 FBc = Ro = 10 KN 925 Similarly for section EMB = 0 Fz=fff RAX Z + P3 x3 + F3 xz = 0 For = -(6+8) = -14XN Q24 from Section 2 Eme = o Rox3+ fer x3 +6200845°x3=0 10+14 - - £24-707) £2 = -24 = -33.946 N Negative Sign indicates foto direction is opposite

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