Question

5. TRUSS Determine the force in each member of the truss shown below. There is a pin at A and a rocker at E. State whether each member is in tension or compression. 150 N 200 N 8 m 125 N 8 m 6 m 6 m

Answer 1

с. ISON - 0 8m RAM 200N A 2 B D Х RAH 125 8m E PRE The members are numbered from | >7 6 m 6 m > Given that it is pin joint at 'A?, and rocker at E Hence at point (A), the finture cannot support any Moments since it is a pin Joint. Moments about (A so taking 150x8 ( clockwise) + 125x6 [ clock cise] + 2004 12 (clockwise] + REXIR (anti abckwise) Nowe EMA = 0

hence 150x8( (w) + læsx66cw+200X12 (<w) TREX12 (CCW) CW que) =) Taking IRRE 150x 8 + 125x6+ 200x12- 12RE 150x 8 + 125x6+200x12 RE 150x8+ 125x6 +200x12 2 12 z 362.5N can And at t have only vectical Reactions as it is a Rockes', hence the reation at point E. point Er 362.50 Now for finding the reactions at 'A' Ź Fx=0 E FY 2 O using & Fx = 0 on the touss. he use RAH+ 150 =0) RAH2 – 150N

implies RAA is in a direction opposite (hence drawn in opposite direction) RAH (leftwards) { Fyzo to Ray + 125 (down) + 2006 down) + RE (up) ے Ray 125(down) + 2006 down) + Reup] down que taking Rav. - [125 +200 - 362-5) 37.5N hence Ray i in down (tve 37.5N) direction. E 2 E Now take Joint Now To | Ty To has a horizontal component which is impossible because there are no other' horizontal forces to balance horizontal component of To RE hence To Should be zelo.

hence the figure changes to ✓ Ty RE hence Ty and RE are egned and opposite 362.50 (downwards) ما کا Ty z 2 so member) in for The memberly) pushes it's ends or end points to be v equilibrium in compression. Tyo 362.50 (in compression) hence member 7 is Jone , The member (6) donut hence it is neither in tension in compression. hence, it has no effect in the touss. nor 14 200 Now lo take Joint (D) TS ↑ Ty the figure of Joxes acting ои Joint of

lom C By с. py the goods Sheorm 8 T4 2oon D BV TS 17 - 362.5nI 6. Applying {fn=0 / { Fy zo each Joints are in equilibrium. Since { Fazo Ty loso + TS2 cos co z 10 6 (from D DCB (hypotenuse lo CD 2 v 62 +6² z lo Tux 6 - TS lo Now for & Fy Tu sino doon) + 200 (down) + T Cup) 20

(tre) taking down С 362.5 Tu sino + 200 Tysino 362.5-200 162.5 z Now sino 8 2 lo 2 TyX8/10 162.5 2 162.5 xio 203.1250 Ty 2 co From O -TS 2 Ty X 6/10 - 203.125X 6 Ty x 6 τς. 6/10 10 121-875N - Ve sign imples (Ty lose) and is are in opposite direction as given by the figure. hence – member pushes each of corners hence is in Compression? T4 203-125 N (compressive) 2

member (5 → at pulls in waeds from inwards from the ends to keep it in esuilibrium, hence is in this stata tensson. Tse 121.875 N ( tensi sion) coming on to joint 'B of T3 This would be the figure Tozo Tz 1 TS 125 Now we know 73 2-125 Z T to hold. Ts for {fnzo and { Fy e a T2=-1s2 - (- 121.875). 121-8250 (implies opposite to Ts] T32 – 1250 which implies oppositi to load 1250

Member (3 pulling from both sides hence is in tension member ③ 2 Tz +1250C Tensile 2. member (2 pulling from both sides, hence is in tension member ③ Ta 2 121.875(Tensile] come to joint, A Now 10kn . TRAV RAH "T 6 takin & Fy Tsino + Ray Tsino a RAY 7 x 8 - 37.5 10 TI - 37.5x10 - 46.875N N

which implies Tisine and Ray are in opposite clirections 1 K member ☺ pulls from either sides, and hence is in tension member I Ti 2 46.875 (tension) like . please leeve if you with answer are happy