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Solution Manual for Concrete Structures, HAR/CDR Edition

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    Solutions Manual to accompany Concrete Structures Mehdi Setareh Robert Darvas

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    CONCRETE STRUCTURES Solutions Manual Table of Contents Chapter One: Reinforced Concrete Technology 1 Chapter Two: Rectangular Beams and One-Way Slabs 6 Chapter Three: Special Topics in Flexure 55 Chapter Four: Shear in Reinforced Concrete Beams 80 Chapter Five: Columns 114 Chapter Six: Floor Systems 150 Chapter Seven: Foundations, and Earth Supporting Walls 158 Chapter Eight: Overview of Prestressed Concrete 232 Chapter Nine: Metric System in Reinforced Concrete Design and Construction 236 1 Study)

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    Problem 1-1 Hydration is a chemical reaction that starts when water is added to cement. It has three stages: setting, hardening, and strength development. The hydration process generates heat. It also continues throughout the life of concrete structures as long as there is free moisture available. Problem 1-2 Since concrete is a construction material which is very strong in compression but weak in tension, it is important for the structural designer to know what the compression capacity of concrete is. The compressive strength of concrete is measured by conducting a “cylinder test”. In this test, compression force is applied gradually on a standard 67x12” concrete cylinder. The stress and strain of the specimen is measured and plotted. The maximum compressive strength is noted asf. Problem 1-3 The air-entraining admixtures arc added to concrete to increase the concrete resistance against the freezing/thawing cycles. As a result, this admixture improves concrete durability.

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    Problem 1-4 The modulus of elasticity relates the strain to the stress in concrete. It can be determined as a result of the cylinder test or the use of the ACI approximate equation. Problem 1-5 The modulus of rupture is the tensile strength of concrete in bending. Problem 1-6 Deformed bars are usually used as the primary steel reinforcement of structural elements. The bars have protrusions on the surface to increase their bondage to concrete. Welded wire reinforcements are thin wires spaced at certain distances in two orthogonal directions and fabricated in large sheets or long rolls. The welded wire reinforcements are usually used where large areas need to be reinforced such as floor slabs and walls. Study

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    Problem 1-7 m2 x 2 ight of beam = —————-=0.22k/ fi weight of 7.000 ft wy =1.0+0.22=1.22k/ fi Ro l2210)_ 2 2 My, =" issp-k 8 w= 1.22] R=61k R=6.1k 2:0" 61k} | vo | ® | | {-6.1k wee © Mowe = 15.3 fk Sketch for Problem 1-7 Problem 1-8 w, =145pcf f!=3500psi F,=57,000)7; = 310003300 _5 355 4; 1,000 J, =7.5y; =7.5v3,500 = 444 psi Study

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    Problem 1-9 w, =120pcf J. =3000psi P w = 30 Ib/ft 6-0" w= 12S =30Ib/ fi 12 12 2 2 My, =P PLO PO) yas sp opis 8 4 8 4 Using Equation (1-3) for sand-lightweight concrete: f= fy = 0.857.517) = 0.85(7.54/3,000 ) = 349 psi 2 2 S, = YA _SO) sim 6 6 Mg =f.S, Mo < My 135+1.5P < (sso) 12 135+1.5P £1,047 P <6081b then P,,, =6081b Study

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    Problem 1-10 f! =4,000psi w=2.1k/ft A beamweight =150) 8.12 =100/b/ fi © 12712 w=2+19 _o usp 1,000 2 M,, = we = 0.26302 f, =7.5f7 =7.5{4,000 = 474 psi 2 2 S, = bd” 302) joi 6 6 474(192) My=fS, =——"2=75%- fi = 12,000 % Mx <M, 0.263%" £7.59 A<537' Study

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    Problem 2-1 f.=4,000psi 1, =60,000psi b=14" A, = 4#9 = 4.00in” Using Method I: (a) d=28” Mg =? From Figure 2-39, use Method I: Step 1. pte 200 46100 bd 14x28 Poin = 0.0033 ( fromTable 42-4) Pas = 0.0207 (from Table A2-3) (Pin = 0.0033) < (p=0.0102) < (Dg, = 0.0207) OK. Step 2. a Asfy __400x60,000 oo, 0.85/,b 0.85x4,000x14 Step 3. e= 2 30 oy B 085 Step 4. La 393 =0212< 3 =0.375 — section is in tension-controlled zone d 28 8 $=090 Step 5. a 5.04 . Mp =¢Af, (¢ - a)- 090x4 00x60 28 SS) 5503.7 k—in =458.6 k- ft Study

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    (b) d=32 Mpg =? Repeat same steps as in part (a) a 5.04 . , My ~gap (1-2) 090x4 00xen{ 32-224 6,367.7 k—in =530.6 k- ft (c) d=36” Mg =? Repeat same steps as in part (a) a 5.04 N Mp =44f, (+-9)- 090x4 00x 36-204 7,231.7 k~in =602.6 k— ft (d) d=40 Mg =? Repeat same steps as in part (a) a 5.04 . My =¢4f, (a-2)- 0. 90a 0x60n{ 0-224) 8,095.7 k~in =674.6 k- ft Comparisons: A linear increase in “d” leads to a proportionate linear increase of Mg % increase of % increase Case “d” over Mg (k-ft) over Case (a) Case (a) Study

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    Using Method II: (a) d=28" Mg =? From Figure 2-40, use Method II: Step 1. pte 400 =0.0102 bd 14x28 Poin =0.0033 ( fromTable A2-4) Prax = 0.0207 ( from Table A2-3) (Prin = 0.0033) < (p=0.0102) < (£4, =0.0207) OK. Step 2. p=0.0102 fy =60,000psi —> Table A2~ 6b — R= 501psi fi =4,000psi Step 3. 2 2 yp, = BER _A9O9CO)_ yo 12,000 12,000 ® d=32" Mg=? Step 1. p= A _ 400 0.00893 bd 14x32 0.0033 < 0.00893 < 0.0207 .". ok Step 2. p=0.0089 Jy =60,000psi —> Table A2—6b—> R=443 psi f{=4,000psi Step 3. Study

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    2 2 py PER _(14)C2 (48) 0 12,000 12,000 (c) d=36" Mg =? Step 1. p= 200 60079 bd 14x36 0.0033 < 0.0079 < 0.0207 .". ok Step 2. p=0.0079 — Table A2-6b — R =397 psi Step 3. 2 2 My=LER (430) (397) _ 0095-4 12,000 12,000 (d d=40 Mg =? Step 1. pt 200 50m bd 14x40 0.0033 <0.0071< 0.0207 .". ok Step 2. p=0.0071— Table A2 6b — R =359 psi Step 3. 2 2 My = bd"R _ (14)(40) (399) _ 670.1% f 12,000 12,000 Study

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    Comparisons: ) % increase of . % increase Case “d” over Mg (k-ft) over Case (a) Case (a) Study

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    Problem 2-2 fl =4,000psi f = 60,000 psi d =36" A, =4#9 =4.00in" @ b=147 Mg =? Use Method I. (Flowchart on Figure 2-39) Step 1. pds 400 0.0079 bd 14x36 Poin = 0.0033 (fromTable A2-4) Pua = 0.0207 (from Table A2-3) (Pin = 0.0033) < (p=0.0079) < (p,q = 0.0207) OK. Step 2. A a= fy __400x60,000 0, 0.85f,b 0.85x4,000x14 Step 3. =a 504 ou A085 Step 4. e395 0.165 3 =0.375 — section is in tension—controlled zone d 36 8 ¢=0.90 ’ Step 5. a 5.04 . M,=¢A fd -5)= 0.90x4.00x60x] 36-=- =7,231.7k—in=602.6 k— ft Study

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    (b) b=16" Mg =? Repeating same steps as in part (a): go A00x60,000 _, 0.85x4,000x16 My=¢4f, (4-4)- 090xs 00036 +41) 7,300 k—in = 6083 k- fi (c) b=18" Mg =? Repeating same steps as in part (a): a= 4.00x60, 000 ~ 3.00" 0.85x4,000x18 Mp =¢4,f, (4-2) 090x400 36-222). 7,352.6 k—in =612.7 k-fi (d b=20" Mpg =? Repeating same steps as in part (a): oo 400x60000 oo, 0.85x4,000x20 Me =¢Af, (+-9)- 0.90x4 eso 36-252). 7,394.8 kin =6162 k-Jfi Comparisons: Due to the increased width, the depth of the equivalent stress block becomes slightly less, leading to a slight increase in Mg % increase of . % increase Case b (in) “b” over Mg (k-ft) over Case (a) Case (a) Study

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    Using Method II: (@ b=14" Mg=? From Figure 2-40, use Method II: Step 1. pt 400 0079 bd 14x36 (Pin =0.0033)< (p= 0.0079) < (p,, = 0.0207) .. OK. Step 2. 0=0.0079 fy =60,000psi — Table A2— 6b — R =397 psi J. =4,000psi Step 3. 2 2 My= bd"R _ (14)(36) C9) _600.3k- fi 12,000 12,000 (b) b=16" Mg="? Step 1. p= 20 40069 bd 16x36 0.0033 < 0.0069 < 0.0207 .". ok Step 2. 0 =0.0069 fy = 60,000 psi — Table A2—6b — R =350 psi 12=4,000 psi Study

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