Question

REVIEW BLEMS oblem solutions All problem sola include FBDs. R7-3. Draw R7-1. Determine Determine the interne normal force, shear force moment at points D and E of the frame. 150 lb -9 ft - R7-4. Draw -3 ft -- 8 ft 4ft - Prob. R7-1

Answer 1

7.1

. C Isolb A = horizontal Reaction force at A Ay avertical reaction force at A By = Reaction force at B D1144 TUH MB - Axe E eft Tey - Ayy 3th 8 ft CF = 8 tan30 take moment at A=0 By x12 = 150 X CF By x12 = 150 x 8 tan 30' [By = 57.735026 lb) Equating vertical forces Ay = By Ay = 57.735026 lb to Equating horizontal forces, Axe = 150lba Consider joint C, 150 lb 30g FAC Fio Equating horizonteel forces FAC Cox 30 =150 - FAC=173.20508 lb forces Fco= - FAC sin":30 Feda - 173.20508 din 30 Equating vertical ve Fcp =-86.60254038 lb

| (8ten 30–1) ft Equaling vestical forces A Fop No= Fro No = -86060254038 lb Normal feroce ald equating honrontal forces [ Voco lb shear force ats Equating vertical forces , TM D = 0 lb ft moment at MD moment Feo ME I Romsft -> uft By Equating horizontal forces Normal force at E INE = 0 lb Equating vestical forces VE + Foot By=0 VE E-By - FD =-57.735026-(-86-2025) TVE = 28.867514 db) Shear force cet E Equaling moment ME = By x 9 + Fco s ME = 57.735026 X9 - 86.202545 TIME = 88.602734 lb ft / moment cet E