Question

Problem 1 Beams (a) - (g) shown below are simply supported and loaded at their ends by two concentrated moments of magnitude PL/10. In addition to the moments, beams (a) – (d) are subject to a uniformly distributed load of magnitude 2P/L, and beams (e) – (h) are subject to a concentrated load at midspan of magnitude P (note the directions of the loads). All beams have length L. Draw the shear and moment diagram for all cases shown. (b). (a) C3

Answer 1

* Given :- Concentrated moments, m=PL 10 Uniformly distributed load , w = 2p Concentrated Load , W =P Length of all beams = DI2 → Ri + R2 = 2P XL = 2P {m, =0 R₂(L) + 4 = 4 + 20 xx x - 4 [R₂ =P] ; [R, =P R. - D PHITT D ITT (SED) 3P (BMD) B.M.cente = = 1 + Plu) - 21+){$) == L + L - P = 3pbe

ITAL > (11111. → B+RS + 2P XL - D > 그 R, + R. - - 2P EM, 0 + +뿌 내는 IR - -9 : LR=- L AN (sep) (SFD) ㅋPL 20 12. B. M.tente = - Ps + (-P)는 + P (는 B, M.centtre = | In 2. | - - - 원 + = - 그는 R - → R+R = ext → 1, + R = 20 |

EM,=0 + 는 +무능 끈 비는 | R(t) + y = PL IR - / | R, = 2P - g - 1 / // 27 - 3 - (SFD) 위 리 (BMD) BMar깥 = p ~ 봉시장 -관광 = 13 13PL so - R > R, FR, + 2P =0 > JR, + R. - - 2P m20 SM), c0 → Ru) + g + P ) () + 우는 = 0 RU+ 2% +PL = 키 = 놓이 2PL + PLO

등에 통해 국 + 75 여 d핑서 TR, OS 화 = ( () + 유 (-) + - - Tue | d . 위 옮 tdt=8

-9111111 9 199 7777773 m (5) $8/1/1/214 IBMD) B.M.centre = P Li + £4) = pole + P4 = 726 -> R,+ R2 + p =0 → | R, + R2 = -P EM-O R2 (1) + PV+14) = path (SED) Py10 A1 (BMD) VP40

--> R, + R2 =P EMso1=무는 1월 4말 R. (t) = ??는 + 일 = S HIHIHUITTITIE. gns) 위으 하 > → R + RL + P = o / R + R - - p SM = o RL)+P(는) - PSP

은 은 (TIS - - - - 10s dc - [as- - - c = (1) 이 Tds -7dC. 7 - 2