Joist CDE loaded as shown Joist CDE Wo= 13 lb/in, L=8ft, E=10^5 psi Joist CDE (DE...

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 [email protected], and use only fost part of equation 1)

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}

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.

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.

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

2 sheer Force Dingdom Bending moment diagram tatti 2اوندا 8

= 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