Question

Consider a 30ft-long barge that carries goods on a river. As shown in the figure below, the barge is subjected to distributed load (3kip/ft) and concentrated load (30kip) due to the goods. Assume that the barge does not move. Draw the shear and bending moment diagrams between two ends of the barge using the graphical method. Make sure that you designate all the important values on the diagrams. 

Answer 1

Answer: As the barge is not moving and placed on a river, so reaction force(buoyancy force) caused by river will be uniformly distributed. So first step will be to find the rate of uniformly distributed load then drawing SFD and BMD. All these calculations are as follows:

Let, reaction (UDL) = We kips/-14 gku/ft ; SF: Shear forre ; 1314; Bending moment gotips — Baz94- 111111112117714_ w kies + E 101 -- Fy = 0 30xwe 10X3+30 Ilokips & w =bo = 2 kipfft - Z +- 10 2okips --- AL 100 kip-ft BM 75k 125k-ft TY Shear porce calculations - Bending moment calculations! for Bc; sfe e ; xe [orio) for Be part: BM : +225;re[0,10) cP: F = -2x+3 0 xe tơ, sổ lớn I boot DE SFD-2x+30+ 3(-15); | BM = 27-30 (-10); xe[e,15 RE[15,25 for De part ? EA: SF = -2x +30 +30 €525,303 BM= 2*2-3062-10)3(-1577 Z NE 15,25] BM2-25 = 625 -4.50 -150 = 25 for EA part 1 BM302-3068-10)-3(x-2013; 26[25,30).

Answer 2

Ans 30 R 3 Klet EBD . Welft For equalibrium 3 x 10 x 5 x 10 q+qFy=0 -3x10-30 +WX30 = 0 w=2 klft o min) = -3X10 (3-10) +2x15x(3-715) +222-15)2 2 20208 630 Nin) = 2X20-3X10 - 30+2 (0-20) = -20+213-20) -3x10x (26-10)+2x20x (JC-10) +2 (36-2012-30(2-20) m(a) = V102) 10K 10K ", SEDLO mone 5! S L20 Kre! 10! 15! 100 kaft 75631 , 25481 25418! 1, .g' 10' 51 .10! BMD Scanned with CamScanner

min) = -3X10 (3-10) +2x15x(3-715) +222-15)2 2 20208 630 Nin) = 2X20-3X10 - 30+2 (0-20) = -20+213-20) -3x10x (26-10)+2x20x (JC-10) +2 (36-2012-30(2-20) m(a) = V102) 10K 10K ", SEDLO mone 5! S L20 Kre! 10! 15! 100 kaft 75631 , 25481 25418! 1, .g' 10' 51 .10! BMD Scanned with CamScanner

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