Question

Using the method of sections, determine the forces in the indicated members of the trusses shown and state if the member is in tension or compression. In addition, summarize the results on a force summation diagram. Notes: o Answers without a force summation diagram will receive a score of zero. Be sure to draw the free-body diagram(s) of the portion(s) of the section(s) used for analysis. Answers without the diagram(s) will receive a score of zero. Show your work. Answers without work shown will receive a score of zero. 1. Members EL and DN. 500 lb 300 TD 300 16 YM 3001b NA .30016 300 lb 300 15 x 30015 150 10 150 lb PA 30 30 - B C D E F G H - 3ft 3f3f3ft 3 ft 3 ft 3ft 3ft

Answer 1

500 lb 300 lb () 30016 - 30016 300 16 300 16 300 lb 150 16 !---* 1501b HE 3f4 34 3f3f3f7 i 3ft 13ft !3 ft Step 1: To find support reaction (Roth ) () {ME = 0 sign -RAX 24 +30 fos 3: X3 + 300 Sin 3* 31035 P300 C330X 6 + 300 Sin 3 x Etano +300 cos 30 x 9 +300 Singox 9 tan3o +5008 12- 30. Sın 30 x 919m 30' + 300 C053015 - 300 Son 30 x 6tan 30+300 C05 30x15 - 3oosin 36 x 3 fan 30 + 300 (0530 x 21+30o cos3ox 24 =0 RA=1289-23 lb . Res 1289.23.16 (7)

growing considering schon 0-0 FRO of left part of on truss and section 100, i for NAV for TA 311 B 3 C R = 1289-23 la for equilinium , ENF =o fon X 9 -300X - 300 X 6 3 Cos 30 1653 fon = 346.11 lb ( Tensile) similarly considering left part of section to 500 Bit Bit C D Bat E PA= 12.89.23 16

(D EME=o -1289.23 x 12 +150 C0530 x 12 +300 205 30X9-300 Sub 30 X 379n 30 + 300 (0536 X 6-300 Sino x 6tan 30 +300 cos 30x3 -300 Sin 38 x 9 tanso . fue sin 66x 12 thn3 =0 : FmL = -1799.04 1h (i) Efy zo 1289.23 -- 150 C05 30 - 300 COS 30300 (0530° -300 60530-500 FML CosGo + Fee Sin Go to - . FEL =. - Goo 1b | FEL = goo lb ( compressive)