Question

Show that the midspan value of
EIS
is $\left ( wb/48 \right )\left ( L^{3}-2Lb^{2}+b^{3} \right )$ for the beam in part (a) of Fig. P-681. Then use this result to find the midspan
EIS
of the loading in part (b) by assuming the loading to extend over two separate interval that start from midspan and adding the results.

Ans. $EI\delta =9280 N\cdot m^{3}$

Answer 1

ww/m Tu BI here ar R a b * lla-b)! " RitR2 = wb {fy=0 [MAIO > (R,XL) = Wb(a+b) = R2 → R, Wblary) and Ri= wb [1-hath) M= moment at distance it from point a Mz Riculº - W42-45+ Water arb)? Po know for deflection in beam using double integration method dy EI duz R, KHy - we-ag² + w <se-a-by² da integration twice the above &qu. I dy heeney? (u-ays us u th-a-by + 4 Kei da 6 EI Y(") + har mai (4- arty! w le-arb) tG <H) +(2. 24 24 Tura) + For constant (,& c2 satisfy boundry condition. » Taro en de (182-a)* + us (4-2-3)* + at uso, yoo atn=l, y so

+ W. (L-a-b) 24h Si referente la casa Eg" of deflection. 4 Jedy(a) = b[1-4 (ato)][r3² - w we Kerast cura-s8 +{mobil (1 - (none) W(2-a) 24 4 w (L-arb) + 24L 24 L as deflection at midle of span. (8) put u=2, a=2. EI 8= wonen (1-1 (4+5)) - ) Wb" (1964); } قدیما 4 twit W. (1.2 b) 1 1 4 24*16L 24 e 18 = { 18-216 487 answer at b) midle span (8.). deflection in beam put w=800 m a=2m, b=3m, L=60, 4= 3m EID = 800x3 x *D* 3 *(1-x) *239 Society (0)*-(scow3xIU x automatis XW"]x3 + 8 00 24x6 » Erd=9280 Noma Answer.

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