Question

IO BEAM DEFECTION - DOUBLE INTEGRATION METHOD () DEVELOP THE ELASTIC CURVE EQUATION FOR THE W14x68 () Slope @c? Ayo (+) DEFVE GICH @c? y By (15)+50(19) Bok =+63.3K By=63.3kN +ZF Ay+63.3-60 ug ay -13,3% Ay=13.3KV

Answer 1

a=4in

L=15in

P=50k

f 15 Let the left support have a vertical reaction RA and the right support have a vertical reaction Ry Sum of forces along the y-axis is equal to zero for static equilibrium: + XF, = 0 RA+Rg - 50 = 0 RARB-500 RA + R = 450kip -(1) Sum of moments about the left support is equal to zero for static equilibrium: + EM=0 R, (15-0) +(50) (19) = 0 R, (15 - 0)+(50) (+19) = 0 RB +63.333kip ----(2) Substitute (2) into (1) RA+Rg = +50 RA 63,333 450 R 13kip

Moment for 0 sxs 15 Take a cut for < < 15: OEM, -0 +(-13.333) (2-0) - M (=) = 0 M(x) = -13.333 M(x) = -13.3332 for 0<=<15

Moment for 15 sxs 19 13.333 kip + Take a cut for 15 <<<19: +OEM,=0 +(-13.333) (3-0) +(63.333) = - 15) - M2) = 0 M() = -949.995 +502 ..M E ) = -949.995 +502 for 15 <<<19

+(3+1)a= Px =”y

M =-Pqx (0<x<L)

so differnetial equation will be

E"*=-PAX

er panta Ely-kepp+Gx+C2 at x = 0, y=0: C2 = 0 at x = L, y=0: 0=-=-PqL+G1 G = Pal

El y +Palt - Pat? (+- (1 11

for x=L+a

For overhang section:

taking

a=b

L=a

a+b=L

a Erine - Poona – aj? +01 -Pb.r3 Ely = PL (x -a) + C1x + C2 60 + 60 x = 0, y = 0, C2 = 0 Pba x = a, y = 0, C1 = x=L Ely = -Pb2° + PLb3 Pabl Ely= 6a 6a 6

b)

$\small \theta=-\frac{pbx^2}{aa}+\frac{PL}{2a}(x-a)^2$

at X=L

$\small \\EI\theta=-\frac{pbL^2}{2a}+\frac{PL}{2a}(b)^2=\frac{PbL}{2a}(L-b)=\frac{PbL}{2} \\\\\theta=\frac{PbL}{2EI}=\frac{50*4*19}{2*29008*722}=7.16*10^{-5} \ rad$

c)by subituing value

$\small \\\\y=\frac{-50*4^2*19}{3*29008*722}=-0.000241918in$