Question

Name ind the deflection at point B in the middle of the simply supported beam. Watch your units; use pounds and inches. 3.60 kips/foot 3.60 kips/foot E = 29,000,000 PSI 8in 4 ft 3 in HINT: 124/2

Answer 1

7.2 kips/4+ 3.6 kips/ 3-6 kips/ft 3.6 3.6 kipsits Kipl -> H+ 7 A & & & 4f+ 4 + RA 3.6 kips/ tu RA Rc 7.2 kipsit Efu zo RA+RC = 1/2 x 4 x3.6 + 1 x4 x3. 6 = 14.4 Kips 33.9 moment about O IMAO REX8 - 4x4x3.6 x ( 13 x ) - 2 x 4x3.6+(4+ 33 x ) = 0 Rc = 6 kips RA= 14.4-6=8.4 Kips By Macaulay as Double Integration method El. dy = me - Rex 43.7.2013 / +3,4(X-472t (2-w - *xx.5x(4+ Y) E2 dl / L = 6X-1.2 x2 / +1.8 (x-4)2 +1.1 (x-412 i 3 x ² ( % +) EI d 2 = 6 - 12 2 +c. / +27 (x-43

By Macaulay / Double Integration Method E de (x from a X2 Rck - x x 7. 2. & 1 / +3.0 (X-413 + 14 (x-4 x. 3 (x-4) (4) - 1 (x-4) 4.9(x-4) (x-4) EI di = 6 x - 1.2 x 1/1.8 (X-432 EI d e xL - 1.2 03 +6, 1 + 12 (x-43 Ely = x3 - 122 x 7 +(, x + 2 / +11 (x-47 cat A (X=o) yao 0 = o-otot (₂ = 0 cat c (X=8) you o= 1x83-1-287+(,8 + (8-494 2) C = -17.0 Ely = x² - 1,2 x ² – 17.6 x 1 + 28 (x-4) for AB x= 4 EL D G = 43 - 14 - 17.6x4 = -32 E = 294 100 PSI PST = 29x10x122 cours 19X10 X12 pound per sq.ft

sin I = I x3 x 83 2 128 tt zin EZ= 128x29x108x122 H" x Pound ft2 32X103 (downward , Do = 32 EI 128x29410° x122 DB = 7.18 x 10t Tuch.