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Desian an economical rectongulay bean size and,select the reinforcing borsvebra Stinply Supported span of 2o feet This beann i to Corry a live lod of 1900 Ibs Jlt and a dead load of 1750 lift. Use a Concrete compresive stress of 3500 put, Mo.3 stirrups, and a Grade 0 ranbrcng Desitn the rechangular beam Codeng deflechoo and bendin 3 9 ing 400 REINFORCED CONCRETE STRUCTURES TABLE 13.1 Properties of Deformed Reinforcing Bars Bar Size Designation Nominal Weight Diameter Cross-Sectional Area lb/ft in in No. 3 No. 4 No. 5 No. 6 No. 7 No. 8 No. 9 No. 10 No. 11 No. 14 No. 18 0.376 0.668 1.043 1.502 2.044 2.670 3.400 4.303 5.313 7.650 13.600 0.375 0.500 0.625 0.750 0.875 1.000 1.128 1.270 1.410 1.693 2.257 0.11 0.20 0.31 0.44 0.60 0.79 1.00 1.27 1.56 2.25 4.00 460 REINFORCED CONCRETE STRUCTURES TABLE 13.12 Minimum Thickness of Slabs or Beams Unless Deflections Are Computeda Minimum Thickness of Slab or Height of Beam Type of Member End Conditions of Span y 40 ksi 60 ksi Solid one-way slabs Simple support One end continuous Both ends continuous Cantilever Simple support One end continuous Both ends continuous Cantilever L/25 L/30 L/35 L/12.5 L/20 L/23 L/26 L/10 L/20 L/24 L/28 L10 L/16 L/18.5 LI21 L/8 Beams or joists し+40. 15501附 LL- 14001b/H 5000 psi SIksi 40 NO. 4 stimp d stirrups 0.5v. yr lawp.owL De Plechan Chart o h-15b to h-ab 30-1.5b 30% ab l6 que ss h-15(13)-a7" bo15" 1) (pe). Fc 5 ksi Go ks maximume 05 Po,0a5 Mr-Mu= "YoukA. I071.llik-Ft 0.40 moksi :14. 085f $5(5ksi) 0 13.65s 32.143 x TRY NEW SIZE* bja, 101 l. 111 k. Ft (lart)一18,055 ina Guess No.g bars der。 Calc ry 1936 400b/e 1.( s Assume P- 0.oiac 3 44161 0.90 m 14.118 Ru- 0.684 ksi bd'.loglll k ft (aP),14,003.3 210 choose b-No. bors arOin 14in Rom)up to 儷儲. Mu 4215 K-p bd 18,on dcalc- 32.417 hi 34917" 35° | revised = O, O 125 Regd As o.o1a5(1335)7.3125n Number ofbars A13120 bor make the bar stne lorger Guess No.ll bars dat1410" Abr 1,5 lein dcal: 32.417 h-3 1.5 0.5" 0.5 (140) 35.1a36" TRY İ8x3G" deale 302w sing No1 borsh 35.191 ng Mn 1041.llI k-F | dachul, 3G-1.5"-0.50.. 0.5 (1 ฯ16")-33a%" 14,11% on lcs ReeJ As# 0.0119(13") (33.215")-1.156in@ check walk IC.GT-lg" CHECK T-C A 156m Grade GD steel fy-o ksi Mn- 109t.I kft nreds to be reiste C0.65 F'cbo T-C 니4qk-085 (5ks) (18), :(o.113 Desigred Mn- 46rk (33.445 -0.14,150 449 k- in-11g Pesign is okay

Answer 1

Given:

Span=20 ft

L.L=1900 lb/ft

D.L=1750 lb/ft

Fc=3.5 ksi

Fy=60 ksi

#3 Stirrups

Step 1:

Calculate the Mu

Wu=1.2(1750)+1.6(1900)

Wu=5140 lb/ft=5.14 kip/ft

M=WuL²/8

M=5.14x(20)²/8

M=257 ft-kips (This is without self weight)

Step 2:

Consider h=0.1L

h=0.1x20ftx12in

h=24 in, consider h=28 in

d=28-1.5-0.375-0.5

d=25.625 in

consider b=h/2=14 in

Now including self weight calculate the B.M

S.W=28x14x150/144=300 lb/ft=0.3kip/ft

Wd=408.33 kip/ft

Wu=1.2(2.108)+1.6(1.9)=5.57

Mu=5.5x400/8=278.5 ft-kips

Step 3:

R_n=278.5x12x1000/(0.9x14x25.625²)

R_n=403.93 psi

$\rho$=0.85f_c/f_y[1-(1-2R_n/0.85fc)^0.5]

$\rho$=0.0074

A=$\rho$bd

A=0.0074x14x25.625

A=2.66 in²

Step 4:

Select No.9 bars of 3 numbers

A_s=3 in²

No of bars=3 to be provided

Now consider 1.5 in cover #3 stirrup and check width

b=2(1.5)+2(3/8)+3(1)+2(1)

b=8.74 in and provided is 14 in

d_actual=28-1.5-0.375-0.5=25.625 in

Step 5:

Check for capacity of beam

a=AsFy/(0.85fcb)

a=3x60/(0.85x3.5x14)

a=4.32 in

M_n=$\phi$AsFy(d-a/2)

M_n=0.9x3x60x(25.625-4.32/2)

M_n=316.76 ft-kips, the design is safe