Assumption:
I am assuming all the loads are factored as the dead loads and live loads are not mentioned
(a) Find out support reactions:
Consider the equilibrium condition of the beam
Fy=0
RA+RB=32+2x30=92 kips
MB=0
RA(30)-32(20)-2(30x15)=0
RA=51.33 kips
RB=40.66 kips
(b) Applied moments:
Separate the beam into two parts a point load beam and UDL and then analyze
M1=Wab/L=32x10x20/30=213.33 ft-kips
M2=wL2/8=2x302/8=225 ft-kips
M=M1+M2=213.33+225=438.33 ft-kips
(c) Cracking moment:
Mcr=frIg/yt
fr=7.5x1.0x(5000)0.5
fr=530.5 psi
Ig=bh3/12=16x343/12=52405.333 in4
Mcr=530.5x52405.33/17
Mcr=136.27 ft-kips<Ma
The cracking moment is less than the applied moment, so the beam has cracked under applied loads
(d) Cracked moment of inertia:
Calculate n value:
Ec=57000(5000)0.5=4030.5 ksi
Es=29000 ksi
Ast=6 in2
n=Es/Ec=7.19 (consider as 7)
Using the transformed area method, calculate the value of the cracked moment of inertia, let us assume the value of neutral axis from the top as x
bx(x/2)=nd'(d-x)
16x(x/2)=7x4x(30-x)
8x2=28(30-x)
8x2+28x-840=0
x=8.64 in
I=bx3/3+nAst(d-x)2
I=16x8.643/3+7x6x(30-8.64)2
I=22601.85 in4
(e) The stresses:
From the above values:
fc=My/I=12x433.8x1000x8.64/(22601.85)=1990 psi (Here the value of y is 8.64 in obtained from above)
fs=nMy/I=7x12x433.8x1000x(21.36)/22601.85=34437 psi (Here the value of y is d-x=30-8.64=21.36 in )
Cracking Moment (Uncracked Concrete Stage) Problem 1) Problem 2.6 (page 55, McCormac and Brown, gth Ed.)...
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