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RE: Enercalc Reinforced Concrete Beam design
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- Subject: RE: Enercalc Reinforced Concrete Beam design
- From: "Brent Koch" <brentk(--nospam--at)tdl.com>
- Date: Wed, 8 Jul 1998 13:04:46 -0700
John: 1) "Which version of the ACI Code is being used?" - I haven't found any specific references to ACI318-95 in their documentation or printouts so you'll have to get that answer straight from the horses mouth (Enercalc). You might mention to them that it would be an improvement to have the Code Reference on the printout or at least as a checkbox option. 2) "Can anyone shed some light on this?" - Below is a "hand calc", of sorts, to check against Enercalc. I input your design data into my SEL ver5.04 to compare results. The Enercalc values are [bracketed] below. My analysis seems to agree with Enercalc's. The 2-#5 T&B with #3 stirrups is Okay based on your problem description. What Enercalc version are you running? Also, you may want to double check your input. Also, you didn't specify the depths of your 2-#5 T&B, so I assumed 1.5" cover to the stirrups and 2.25" to the cg of the rebar layers from face of concrete T & B. In any event the Enercalc folks will want you to fax them a copy of your calculation printout & a complete problem description. Would appreciate a follow-up post if your communications with Enercalc turns up anything interesting. BTW - where is pwcsd.nosc.mil ? Regards, Brent Koch FLEXURE ^^^^^^^ DLbm(8"x12"@150pcf) = 100 plf ADL (given) = 1064 plf ---------------------------------- TOTAL Unif DL (Service) = 1164 plf x 1.4 ---------- TOTAL Ultimate DL = 1630 plf For fixed/fixed: Mu(-) = Factored Support Moment = -(1630*12^2)/12 = -19560 lb-ft [Enercalc = -19.56 k-ft] Mu(+) = Factored Misdpan Moment = (1630*12^2)/24 = 9780 lb-ft [Enercalc = 9.78 k-ft] Given: b= 8" , h=12" , f'c=3000 psi, Fy=60,000 psi Top As = 0.62 in^2, dtop = 2.25" Bot As = 0.62 in^2, dbot = 12"-2.25" = 9.75" Neglecting compression steel to simplify analysis: T = C = As*Fy = 0.62*60,000 = 37,200 lbs a = C/(f'c*b*0.85) = 37,200/(3,000*8*0.85) = 1.82" jd = d-a/2 = 9.75-1.82/2 = 8.84" phi Mn = phi*T*jd = 0.9*37,200*8.84/12 = 24,660 lb-ft [Enercalc = 24.63 k-ft] SHEAR ^^^^^ max Vu = 1630*12/2 = 9780 lbs [Enercalc = 9.78 k] phi Vc = phi*b*d*2*sqrt(f'c) = 0.85*8*9.75*2*sqrt(3000) = 7260 lbs Requ'd phi Vs = 9780 - 7260 = 2520 lbs min spacing governs (by inspection)--> provide #3 @ d/2 = 4.875" min o/c for #3 @ 4" Av/s = (2*0.11)/4 = 0.055 in^2/in = 0.66 in^2/ft Provided phi Vs = phi*(Av/s)*d*Fy = 0.85*0.055*9.75*60,000 = 27,350 lbs min Av/s = 50*b/fy = 0.007 in^2/in = 0.084 in^2/ft >-----Original Message----- >From: John Armijo [SMTP:421jarmi(--nospam--at)pwcsd.nosc.mil] >Sent: July 07, 1998 10:56 AM >To: seaoc(--nospam--at)seaoc.org >Subject: Enercalc Reinforced Concrete Beam design > >I am designing a reinforced concrete lintel over a 12'-0" masonry >opening. I have input 2 # 5 Top and Bottom of a 8" wide by 12" deep >beam. End conditions are Fixed-Fixed, Seismic Zone 4, 150PCF concrete >weight, uniform dead load of 1.064 KLF, and #3 ties. All other input values >are the defaults. > >I have two questions: > >1. Which version of the ACI Code is being used? > >2. For the input above I get Maximum Moment, Mu= -19.56 K-ft and >Allowable Moment, Mn/phi= 16.48 K-ft and a note saying, "Beam Design >OK". Granted a positive number is larger than a negative number, but my >concern is that the absolute value, | -19.56 | is greater than the absolute >value of | 16.48 | and that the beam design is NOT OK. > >Can anyone shed some light on this? >
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