beam is continuous over two equal 22 ft spans simply A reinf

beam is continuous over two equal 22 ft spans, simply A reinforced concrete supported at the two exterior supports, and fully continuous at the interior support. Concrete cross-sectional dimensions are b 10 in., h 22 in., and d 19.5 in, for both positive and negative bending regions. Positive rein- forcement in each span consists of two No. 9 (No. 29) bars, and negative reinforcement at the interior support is made up of three No. 10 (No. 32) bars. No compression steel is used. Material strengths are, 60,000 psi and fc = 5000 psi. The beam will carry a service live load, applied early in the life of the member, of 1800 lb/ft distributed uniformly over both spans; 20 percent of this load will be sustained more or less permanently, while the rest is intermittent. The total service dead load is 1000 lb/ft including self weight.

Solution

Span=22 feet

Fy=60000psi

Fc=5000 psi

n=8

y=11 in

Ig=bh³/12

Live load=1800 lb/feet

Dead load=1800 lb/feet

The Live load retained=1800x0.8=1440 lb/feet ( for long term deflection)

Mcr=7.5(Fc)^0.5Ig /(y)

Mcr=7.5x5000^0.5x8873.33/11

Mcr=35.65 K-ft

Ec=57000(5000)^0.5=4.03x10⁶psi

Load=1.8K/ft

The permanent load=1.8x0.8=1.44 k/ft

Calculation of positive and negative cracking moment of inertia

Ig=8873.33 in⁴

Ie=(Mcr/Ma)³ I_g+[1-(Mcr/Ma)³]I_cr

for positive cracking of inertia

Mcr=35.65 k-feet

obtaining from the transformed area

10x²/2=8*2*(2.5-x)

x=1.64 inches

I=1/3*10*1.64³+16x(19.5-1.64)²

I_cr=5107.676 in⁴

Ma=3.6x22²/12=145 k-feet

I_e=5168.86 In⁴

calculation of moment of inertia in the negative region

I_{g=8873.33in⁴}

Ma=72.5 k-feet

I=6572.4 in⁴

Ie=0.7(5186.6)+0.15(6572.4+6572.4)

Ie=5601.62 in⁴

The deflection is given by

5wL⁴/384EI_e

=5x1.44x22⁴/(384X5601.2X3122)

=^{0.25 inches}

6.5

short term deflection:

Ig=8873.33in4

Mcr=35.6 k-feet

Ma=1k/ftx22x22/8=60.5 kft

Icr=5168.86 in⁴

delection=0.125 inches

For full load deflection

Deflection=0.52 mm

Initial deflection for full live load=0.4 inches

the deflection for 20% of live load

is given by

M=(DL+0.7LL)LxL/8

deflection=0.99 inches

final deflection=0.99-0.4=0.76 inches