Question

Based on tributary load analysis, the dead and live loads, wd and wų, respectively, acting on a beam in a vertical load resisting system are shown below. The concrete is normalweight with compressive strength f=5000 psi. Note that the given dead load includes the self-weight of the beam and slab. Section wp=1.2 kip/ft, wu=1.5 kip/ft be-45 inch 5 inch 1 30 inch B A 10 inch In=10 ft 1. Design and detail the beam for positive flexure at section A. Calculate the flexural demand using the 1.2D + 1.6L LRFD load combination. Recall that for flexure, M = w1A/8 gives the highest possible demand in a beam span. Use a single layer of Grade 60 longitudinal reinforcing steel. Assume 1.5 inch clear cover and #4 transverse reinforcement. Show that your design satisfies OMn > Mu and that bar spacing requirements are satisfied. You can neglect the effective flange width, be, of the slab (shown in the figure) in your strength calculations. Draw a figure of the cross-section showing the longitudinal reinforcement and dimen- sions of clear cover and bar spacing. 2. Design and detail the beam for negative flexure at section B. Calculate the flexural demand using the 1.2D + 1.6L LRFD load combination. Recall that for flexure, M = w12/12 gives the highest possible negative flexural demand in a beam span with fixed ends. Use a single layer of Grade 60 longitudinal reinforcing steel. Assume 1.5 inch clear cover and #4 transverse reinforcement. Show that your design satisfies oMn > Mu and that bar spacing requirements are satisfied. Draw a figure of the cross-section showing the longitudinal reinforcement and dimen- sions of clear cover and bar spacing.

Answer 1

Section Given data h = 30" bw=10"= b be=45 inch 5 inch In = 10 ft Factored load 30 inch W DL=1.2 k/ft WlL=1.5k/ft ASCE 7 LRFD Load combination 10 inch Factored dead load =1.2w pz +1.6*w, LL wu=1.2*1.2 +1.6*1.5 = 3.84 k/ft Material properties fc= 5 ksi f'y= 60 ksi

1) Design for postive flexture at A Maximum moment in the center of span wu*12 = = Mu 8 3.84*102 Mu = 48 k - ft 8 Mu Mnrequired 6 Assume o as 0.9 (Tension controlled ) 48 Mnrequired = 53.33k - ft 0.9 = 640 k - in Effective depth of beam clear cover = 1.5" Assume #7 bars arranged in one layer) dia of #7 bars db =0.875" dia of #4 bars (transverse reinforcement) db =0.5" d = h-clear cover -0.5* db-db =30-1.5-0.5*0.875-0.5 = 27.56" shear

1 1 Required steel ratio 0.85*fc 2* Mnrequired P fy 0.85* fc*b*d? 0.85*5 2*640 р *(1-1- -) = 0.001418 60 0.85*5*10*27.562 required rebar area As =p*b*d = 0.001418*10* 27.56 = 0.391 in? 0 As(min) ACI 318 specify minimum area of rebar to be provided bw*d max(3* fc ,200) fy bw = b = 12" max(3* fc ,200 )= max(3* 5000 ,200) = max(212.13 ,200) = 212.13 212.13* 27.56*10 A = 0.974 in? S(min) 60,000 As required = 0.391 <0.974 in? Hence provide As min

for # 7 bar Area of bar = 0.6 in? 0.974 no of bar = 1.62 bar 0.6 hence 2 no bars of # 7 bars As provided = 2 *0.6 = 1.2 in? 0 check for Spacing of rebar ACI 318 25.2 clear spacing shall be at least the greatest of (1 in., db, and (4/3)/ * dagg.) greatest of(1,0.875") =1" clear spacing= 10-2*1.5-2*0.875-2*0.5 = 4.25">1" 1

45" Hanger 5" #4 stirrups 30" 4.25" 1.5" #7 10"

Elimit check 1)strain in the steel rebar ACI 318 specify limiting minimum strain value in the rebar fy +0.003 Es where Es = 29000 ksi 60 E limit +0.003 29000 = 0.00507 for our section depth of stress block As* fy 0.85* fc*b 1.2 * 60 = 1.694" 0.85*5*10 nuetral axis depth a Bi Bi = 0.8 for f'c = 5 ksi 1.694 = 2.118" 0.8 a = C=

&= 0.003 * strain in the rebar (27.56 -2.118) 2.118 = 0.036 >0.00507 hence tension controlled member $=0.9 Assumption is valid Capacity of section $Mn = $* fy * As *(d- 5-1.6942) =0.9*60*1.2(27.56-1.694 = 1730.905 k - in = 144.242 k-ft Mu = 48k - ft :: Mn > Mu

2)

Design for negative flexture at section B wu* 12 Mu= Mu = 12 3.84*102 = 32 k-ft 12 Mu Mnrequired 6 Assume o as 0.9 (Tension controlled ) 32 Mnrequired = 35.556k-ft 0.9 = 426.667 k - in

shear Effective depth of beam Assume #7 bars arranged in one layer) dia of #7 bars db =0.875" dia of #4 bars (transverse stirrups)db = 0.5" clear cover = 1.5" Effective depth d = h - clear cover -0.5 * db-db, = 30-1.5-0.5*0.875-0.5 = 27.56" Required steel ratio 0.85* fc 2* Mnrequired P fy 0.85* fc*b*d? 0.85*5 2* 426.667 р *(1-1- -) = 0.001 60 0.85*5*10* 27.562 required rebar area As =p*b*d =0.001*10*27.56 = 0.26 in? As min will govern ** 1 =

As(min) =0.974 in( From 1) As required = 0.26 <0.974 in? Hence provide As min for # 7 bar Area of bar = 0.6 in? 0.974 no of bar -=1.62 bar 0.6 hence 2 no bars of # 7 As provided = 2*0.6 = 1.2 in? 0 check for Spacing of rebar ACI 318 25.2 clear spacing shall be at least the greatest of (1 in., db, and (4/3)/ * dagg.) greatest of(1,0.875")=1" clear spacing = 10-2*1.5-2*0.875-2*0.5 = 4.25"> 1" 1

45" #7 7" 4.25" #4 stirrups 30" Hanger 10"

a = strain in the steel rebar(Same calculation as 1) Depth of stress block As* fy 0.85* fc*b 1.2*60 = 1.694" 0.85*5*10 nuetral axis depth from bottom a C= Bi Bi = 0.8 for f'c = 5 ksi 1.694 = 2.118" 0.8 strain in the rebar (27.56 -2.118) = 0.003* 2.118 = 0.036 >&limit=0.005070 hence tension controlled member $=0.9 Assumption is valid

Capacity of section * As*(0-2 زم Mn = 6 * fy * As*(d. 2 =0.9*60*1.2(27.56-1.694 =1730.905 k - in = 144.242 k-ft Mu = 32k-ft ..Mn > Mu

Note

beam is lightly loaded and span is relatively small

so minimum rebar required governed the design for both section

contribution of flange is not taken for both design