Question

joist CDE loaded as shown

Joist CDE (DE overhangs) 1.) Find deflection curve equation bw C to D and between D and E using 2nd order integration Wo y E с X 42 1441 2.) determine deflection Value and Slope at roller (E, I, Wo; L) 3.) shear force & bending Moment diajam 4. determine stress @.3-in. principle and 44 to the right of C 2) determine & show Stresses angle.

Joist CDE

Wo= 13 lb/in, L=8ft, E=10^5 psi

Answer 1

X Solution - * x T wo D С E Rc 42 RD 42 X 4 х (De overhangs) Using 2nd onder Integration, Taking section at distance x from 'c' to find Deflection Cerove equation. o support Reactions Emc=0 & tve G-ve Ro XL Wo X L XL 2 2 Rp = WL Efy = 0, i tre dove Rc + Rb = wol

Rc = wol - Wool Rc = y According to above calculations, RC = 0, Hence, Taking section at distance u form E to find deflection Carove equation. m = EI d²y da2 where, ma Bending moment at section xx from E. Wo:21:22 (WOL) (8-5) X* m = EI. d2y dx2 2 Integrating ist w.g.tox 3 2 EL: (wol) (-ų) Wo.33 +6,1 + dy da ( 2 = 6 2 to x Integrating and win uzito 3 (x-2 (WOL) EL 4 = Wo. 24 (2} + C, 21 7 6 24 2 whese, C, and (2 = Integration constants. To find , $ (2, Use boundary conditions, at x = { y = 0, Put this in equation@, and use only fost part of equation 1)

. O = Wo o = Wo() +C를 + 2 그래 도 || Wo XL4 ci 디 2 21 x 1 .', 2 3 1 Wo4 38 4 C,느 륵 2 Similarly. ) At x=L, y = 0, Use whole equatione number 8, EL.9 = WAY + C, A + C - (No) (-) 그니 G 4 0) . 0 - W (-) + C, L - eol - 는 f(error) 그니 384 1) (늘 6 67Cm = 0. +C,는 2 wo4 324 (NoL) 13 X 2 CO 6 Wol4 . ㅇ wa4 Wol4 + C, 24 2 384 18 . 0 = 0.0 32 OLY + Cry. : C = -60032 0 14 - - C03EL 03 L 2 C: -0-0364 W}

+ 0.018 2 Wol4 -Wo24 Put this value in equation , C2 = (-0.0364 Wols) -WOL4 + 0.0364 Wol4 384 - Q 384 2 C - - C2 Wol4 384 6 Wol4 384 3 wol4 Wol4 C2 3 - C2 64 192 G wol4 64 - 2 2 Put, C, and C2 Values in equations of 2 - 0.0364 Wolf (WOL)(x-u? El. dy Wo X3 2 dre 6 (14) This is slope equation similarly, wox4 .000364 wo L².2 + wol4 EI Y 64 24 3 - (WOL) (2-5) 6 This is deflection equation.

© To find slope and deflection at D. Put x = { in equation 0 and 5 first part of equation upto bracket. and use only a) El. dy wor3 3 -0.0364 Wol da 6 3 :, El. dy WO (5) = 0.0364 Wo () da G :, Wol² = 0.0364 Wol? 48 2 3 3 -2 wol 1.7472 wol il 48 8 2 : En dy -(WOL) (2.633 X 16-3) da 3 -3) شاہین x ::.. dy EL. = (2.63 2.633 X 10 da = = 8X12 = 96 inch Given Wo 13 lb) inch, L=8 Ft E - 105 psi -3 Wol² dy da at D = Slope at D = 2.633 x 10 EL.

6) EI.Y wo 24 0.0364 wol²x + Wola 64 24 Put a L 2 :, EL Y = wo woll (1/2 4 0.0364 WolóxL+ 2 64 24 4 Wol4 » 0.0364 wol wol4 t 244 x 16 2 64 WOL4 0.0182 wol4 EL. Y = Wo24 64 + 384 O Y = 0 EL. y Wo D :. Thus, Hege proved that at support, Deflection is zero. shear Force Dlags am calculations Rc = 0, RD = WOL sign convention dl 3 C 2 RD = RC the uld 느 2 -ye = SF = 0, 2 s. For won - = - WOL worth wol 2 a sfoL - I sFCR WOL - WOL = 0

© Bending Moment Calculations - 6/? -ve B.ME = 0 2 B.MD - Note = 느 4 wol ul 8 Bm 'C - 0 Answer = (3 wo D E с MT 느 2 WOL X D E С. WOL . 2 sheer Force Dingdom Bending moment diagram tatti 2اوندا 8

Answers slope equation foom Etop As we know, During calculations, Rc = 0, Hence we need to take equations foom E. .y Х 3 = da 6 El. dy Wo23 -0.0364 WOL3 slope equation from o to c. EI. dy -0.0364 W3C-WOL (3-5) 2 3 را Woa dre 6 2 from Eto D, @ Deflection equation ELY. = wor4 24 0364 wo L²x + wolk and from D to C, 24 El. Y wo 0.036460023X + Wo24 64 24 (woL) (x - 1 2 3 6 : © slope at Roller D = 2.633 x 10 -3 Wol3 Se EL Deflection at Roller D = Zero u