HW%208%20Solution - CE4401 Fall 2009 Homework#8 Due Wed Nov...

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Unformatted text preview: CE4401 Fall 2009 Homework #8 Due Wed. Nov. 11 For all problems (for this homework and future homeworks) assume a #3 tie and max agg size of 3/4” unless otherwise stated. You may make any other assumptions necessary, but be sure to state them in your solutions 1. Make a list of all the references you have seen in class or doing homework to the ACI code this list should include the classes through Wed Nov. 4 and this homework assignment. Include the code section, code page number, the topic, and a brief summary of what the code says in that section if possible ( i.e. if you can summarize it in a few words otherwise leave this column blank). Do not copy word for word out of the code. For Example code code Topic I summary section ae ' must be deformed ' 10.2.4 stress-strain assumtions for steel elastic-erfectl lastic 2. A 20 ft long simply supported rectangular concrete beam supports a uniform service dead load consisting of its own weight plus 0.5 kips/ft and a uniform service live load of 2.3 kips/ft. The width of the cross section is 14 inches. The overall depth of the cross section is 22 inches. The effective depth of the cross section (distance from extreme compression fiber to the centroid of 19-6" the steel) is 19.6 inches. The beam is reinforced with 3 No. 9 bars. The concrete strength is 4000 psi and the yield strength of the reinforcement is 60,000 psi. The concrete is normal-weight concrete. Compute the design 9 C . flexural capacity of the beam (oMn). Is this beam safe? On a sketch of the cross section at midspan, show the location of the compression zone. What is the strain in the steel at ultimate? 3. a) Compare ¢Mn for singly reinforced rectangular beams having the following properties: No 22 3 No. 8 b) Taking beam 1 as the reference point, discuss the effects of changing As,fy,fc', b, and d on ¢Mn, What is the most effective way of increasing ¢Mn? What is the least effective way? Look at the % change in the variable being changed and the resulting % change in ¢Mn. 4. A 30 ft span simply supported beam has a rectangular cross section with b = l4 inches, and h = 32 inches. The cover is 1.5 inches and the beam uses a No. 3 tie. The beam is made from normal— Weight 4000 psi concrete and has 9 No. 8 Grade 60 bars (in three rows). This beam supports its own dead load plus a uniform service superimposed dead load of 0.5 k/ft. Compute the maximum service live load that the beam can support. CE 4401 Problem 2: Given 11/11/2009 HW #8 KEY 2.1 A 20 ft long simply supported normal weight rectangular concrete beam supports a uniform service dead load consisting of its own weight plus 0.5 k/ft and a uniform service live load of 2.3k/ft. The width is 14 inches, the height is 22 inches, and the effective depth is 19.6 inches. The 4000 psi beam is reinforced with 3 grade 60 #9 bars. Compute the design flexural capacity ¢Mn. ls this beam safe? Sketch the location of the compression zone on the cross section at midspan. What is the strain in the steel at ultimate? Beam length L :2 20ft Beam width b 2: 14-111 Beam height h := 22~in Dead load WD 2: 0.5-E9 ft Live load wL :: 2.3—ij ft Effective depth d 2: 19.6-in Concrete strength fc z: 4000-psi Rebar yield stress fy :: 60-ksi Total steel area AS :2 3-1.00-in2 = 3~in2 Determine factored moment Self weight of concrete «(C :: iSOpcf . kip Self weight w := «b~h=0.321-—— S “{C kip Factored load Wu := 1.2-(wD + wS) + 1.6-wL = 4.665? t wu-L2 Factored moment Mu : = 233‘kip-ft 8 Determine moment desi n stren th A f Depth of compression block a ;= A 2 378411 0.85-fc-b Assume strength (in 2: 0.9 reduction factor Nominal moment strength 7 Mn := AS-fy-[d — 3) = 266~kip~ft Design moment strength ¢~Mn = 239-kip-ft Check moment capacity (13an Mu Check := if [ < 1 , "This beam is not safe for moment" ,"This beam is safe for moment") CE 4401 11/11/2009 HW #8 KEY 2.2 Problem 2: (Continued) Check = "This beam is safe for moment" Compute strain in steel at ultimate Maximum strain in concrete ecu := 0003 Factor 01 := 0.85 Depth to neutral axis 0 := —a— : 4.45-in B1 . . . Ecu‘m — 0 Strain in steel at ultimate at z: —-————— = 0.010 The strain in the steel at ultimate (>0.005) validates the assumption of it) = 0.9 The design moment strength ¢Mn is 239 kip-ft which is safe for our factored moment of 233 kip-ft. The strain in steel at ultimate is 0.010 and a sketch of the copression block at midspan is shown below. CE 4401 11/11/2009 HW #8 KEY 3.1 Problem 3: a) Compute ¢Mn for singly reinforced rectangular beams having the following properties b) Taking beam 1 as the reference point, discuss the effects of changing AS, fy, fC, b and d on ¢Mn. What is the most effective way of increasing (Wm, what is the least? Look at percent change in the variable versus the percent change in (Mi/in. Beam 1 Width b1 2: 14-in Effective depth d1 :: 20-in . 2 . 2 Steel area A81 := 3-0.79~1n = 2.37m Concrete strength f01 :2 4000-psi Steel yield strength fyl 2: 60000~psi Factor 011 := 0.85 Ultimate concrete strain 5011:: 0.003 . Asl‘ y1 Depth of compressron block a1 := ————————— = 2.99 in 0.85-f -b C] 1 Design moment strength at _ assuming q; = 0.9 ¢Mn1'" O‘g'Asl'fy1'(d1 ‘ ‘3 _ 197191”: a1 Depth to neutral axis c1 := —— = 3.51 in 611 Strain in steel at ultimate, ._ €cu‘(d1 _ C1) _ 0 014 ¢=0.9 if greater than 0.005 Etl ’_ c1 ‘ ' Beam 2 A -f 2 1 A52 2: 34.004112 = 3m2 S y = 3.78 in a ‘= —— 2 ‘ 0.85-fC1-b1 a2 CE 4401 Problem 3: (Continued) a 2 c2 2: — = 4.45 in 511 Percent change in variabie Percent change in ¢Mn Ratio to determine efficiency Beam 3 fy3 2: 4000O-psi a C3 — —3' = 1n 511 2-: - d — c 1 3 EB = C“( ) = 0023 C3 Beam 4 fc4 := SOOO-psi I: 0 a4 C4 = —- = m .814 M 4 Rati04 := m = 0.065 %Variab1e4 1 1/1 1/2009 HW #8 KEY 5:2 := -~————— : 0.010 A — A 2 1 %Variab162 := (5—5) 2 27 % A51 43M — ¢M %M0mcnt2 := L—EL—n—l) = 24% ¢Mn1 M RatioZ z: W = 0.898 %Variab162 Asl‘fy3 = ——-— =1.99in 0.85-2.14:1 a3: a3 f _. %Variable3 := M = —33 % fyl ¢Mn3 " ¢Mn1 %Moment3 :2 ————--— : —32 % ¢Mn1 M t ' Ratio3 := W = 0.946 %Variable3 Asl‘ y1 . a4 := —---—-- = 2.39 In 0.85-fc4~b1 a 4 . d _ em := M = 0.017 C4 f - f %Variable4 := (—C4—C—1) = 25 % fcl 43M 4" (MM %M0ment4 := M = 2 % Cme 3.2 CE 4401 11/11/2009 HW #8 KEY 3.3 Problem 3: (Continued) Beam 5 A . b5 := 16-111 215:: ——S—1—¥l— = 2.61111 0.851fC1-b5 a5 211 b —b 05 := --- = 3.51 in %Variab165 := M = 14% b 811 1 5 . d — c ¢M — 111M 1 5 1 s6 := flL—w—Z = 0.014 %M0ment5 :2 LEE—IL) = 1% C1 ¢Mn1 7M 15 RatioS 2: 1—931-63— 2 0.071 %Variab1€5 Beam 6 d = 22-111 a1 6 ¢Mn6 2: 0.9-Asl- yl- d6 — —2— = 219kip-ft a1 a - d6 — c 116:: —- = 3.51111 em := ill—~32 = 0.016 311 ‘36 d —d ¢M —¢M 1 1 %Variable6 := £6—1 = 10% %Momcnt6 :2 (—né—n) = 11% d1 ch111 M Ratio6 := %°—mem6 = 1.081 %Va11‘able6 This is the answer for part a with beams listed in numerical order with beam 1 at the top and beam 6 at the bottom: Answer for part b ‘ ll of the variables in this parametric study are directly related to ¢Mn meaning that an ncrease in the parameter results in an increase in ¢Mn. AS, fy and d are all relatively ffective in increasing (11Mn while 1C and b are relatively ineffective at increasing 0M”. l rder from most effective to least effective in changing ¢Mn: d, fy, AS, b and fc. CE 4401 11/11/2009 HW #8 KEY 4.1 Problem 4: A 30 ft span simply supported beam has a rectangular cross section with b = 14 inches and h = 32 inches. Cover is 1.5 inches and #3 tie is used. Beam is normal weight concrete and grade 60 rebar is 9 #8 bars in 3 rows. The beam supports its own dead load plus a uniform dead load of 0.5 k/tt. Compute the service maximum live load that the beam can support. General Beam length L = 30ft Beam width b = 14-in Beam height h = 32-m Dead load wD 2: 0 5-3“—2 ft Tie diameter due — 0375-111 Rebar diameter db = LOO-in Clear cover cc = 1.5-in Self weight of concrete «(c :: 150-pcf Concrete strength fC := 4000~psi Rebar yield stress f := 60-ksi Total steel area Vertical clear distance y AS := 9-0.79-in2 = 7.11-in2 between rows of rebar S ;= Lin Steel modulus of elasticity E :2 29000.ij fy Yield strain of steel 5y := —— = 000207 E Maximum strain in concrete gcu ;= 0003 Factor 61 := 0.85 Determine if steel has ielded g l rm?) AS-fy V Depth of compression block a = = 8.96.1n 0.85-fC-b Depth to neutral axis 0 z: i = 10.54in 51 Effective depth _ for bottom row .= h _ CC — dtie " -— 111 Effective depth for middle row Effective depth for top row (12:: (11 —db ~s=27.6in I: "db —s=25.6in CE 4401 11/11/2009 HW #8 KEY 42 Problem 4: (Continued) Steel strain for Eculdl — c) bottom row 51 1: w = 0.00543 C Steel strain for €cu'(d2 ' C) middle row 52 - = 0.00486 C Steel strain for 6cu'(d3 _ C) bottom row ' E3 ' = 0.00429 0 We see that all of the steel has yielded so our value for a using the steel yield stress is correct. We can compute the moment capacity using the average of the three cl values which is d2. Determine moment design strength Assume strength (1):: 0.9 reduction factor . a _ Nominal moment strength Mn 2: AS-fy-[dz — = 823-hp-ft Design moment strength ' (ta-Mn = 740~kip~ft The strain in the extreme tension steel at ultimate (51>0. 005) validates the assumption of q) = 0.9 Determine maximum service live load Maximum factored h 8-¢-Mn _ 658 kip uniform load Wumax -— 2 — . L . kip Self weight WS :: fidbh 2 0.467.__ Maximum service ‘ . live load and round w z: ——-———-—-———— = 3.39.— - L 1 6 ft down conservatively - ...
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This note was uploaded on 06/09/2010 for the course CIVIL 4401 taught by Professor Shield during the Fall '09 term at University of Minnesota Crookston.

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HW%208%20Solution - CE4401 Fall 2009 Homework#8 Due Wed Nov...

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