Errata as of February 3, 2020 for ACI SP-17M(14) The Reinforced Concrete Design Handbook - A Companion to ACI 318M-14 (Metric Edition) PDF

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ACI SP-17M(14) V. 1 The Reinforced Concrete Design Manual Errata may already be included in the print version you own

ERRATA as of February 3, 2020 p. 60, One-way Slab Example 2, Step 11, third column, equation revised as: ACI 318M-14 Discussion Calculation Step 11: Minimum and maximum spacing of flexural reinforcement 7.7.2.3 The maximum spacing of deformed 3(175 230 mm) = 525 690 mm > 450 mm reinforcement is the lesser of 3h and Therefore, Section 24.3.2 controls; 300 mm 450 mm p. 72, One-way Slab Example 3, Step 8, third column, m2 should be mm2 ACI 318M-14 Discussion Calculation Step 8: Calculate required As 22.2.2.4.1 Note that the effective depth is 123 mm at critical locations, except at the φMn=(0.9)(116.5mm2/m2)(1395MPa)(125mm-4.64/2 mm) exterior joint. Therefore, the Code permits a minimum of d of 0.8h, or 120 mm For 20 mm cover, φMn = φAps fps ( d − a / 2 )

p. 202, Beam Example 4, Step 7, third column, kN should be kN·m Discussion Calculation ACI 318M-14 Step 7: Torsion design 9.5.4.1 Check if torsion can be ignored; does Tu@d = 22.5 kN·m >φTth=20.3 kN·m Tu @ d < fTth

p. 223, Beam Example 5, Step 5, first column, should be revised as: ACI 318M-14 Discussion Calculation Step 5: Moment design Re-evaluating the positive negative tension reinforcement at girder midspansupport:

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p. 228, Beam Example 5, Step 6, third column, Eq. should replace 450 mm with 600 mm as shown: ACI 318M-14 Discussion Calculation Step 6: Shear design Specified shear reinforcement must be 9.6.3.3 Av,min 450 mm ≥ 0.062 35MPa = 0.393 mm 2 / m at least: s 420 MPa 0.062 f c'

bw b and 0.35 w fyt f yt

Av,min 600 mm ≥ 0.062 35MPa = 0.393 mm 2 / m s 420 MPa

Because fc ' > 31 MPa, Eq. (1) controls: p. 254, Beam Example 6, Step 6, third column, Eq.(c), fsy should be fpy as shown: ACI 318M-14 Discussion Calculation Step 6: Design strength (a) Flexure 20.3.2.4.1 For unbonded tendons…..the stress in post-tensioned tendons at nominal (a) fsypy=0.9fpu = (0.9)(1860 MPa) = 1674 MPa flexural strength is the least of: (a) f se + 70 + f c' / (100r p ) (b) f se + 420 (c) f py p. 350, Rigid Diaphragm Example 3, Step 19, should be revised as shown: Step 19: Discussion When center of mass and center of rigidity do not coincide, the lateral force in resisting systems perpendicular to the direction of the seismic force are subjected to shear forces as calculated in this example: (W1 (121 kN), W4 (47 kN), and W5 (74 kN)) for a seismic force acting in the N-S direction and….. p. 500, Retaining Walls Example 1, Step 4, first column, insert text as shown below: ACI 318M-14 Discussion Calculation Step 4: Wall design Equation (9.6.1.2(a)) controls, ….. 21.2.2 Use No. 13 at 300 mm on center each to resist…..

If As provided at every section is at least one-third greater than As required by analysis, 9.6.1.1 and 9.6.1.2 need not be satisfied. Check if the tension controlled assumption….

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p. 511, Retaining Walls Example 2, Step 8, first column, insert text as shown below: Discussion Calculation ACI 318M-14 Step 8: Heel design 7.5.1.1 Shear The base heel is designed for shear….not be linear as assumed (refer to Fig. E2.6) Note: In the 2019 edition of ACI 318 Section 13.3.6.3 was added, which sets the critical section for shear and flexure at the interface of the stem and footing. Checking the calculations: Stem: fVn = 221.8 kN > Vu = 96.7 kN Heel: fVn = 204 kN < Vu = 231 kN revising the depth to 0.45 m, fVn = 292 kN > Vu = 274 kN OK

p. 512, Retaining Walls Example 2, Step 8, third column, insert equation as shown below: ACI 318M-14 Discussion Calculation Step 8: Heel design Heel: fVn = 292 kN Design shear strength: Factored shear Vu = 212 kN fVn > Vu fVn = 204 kN Vu = 274kN

p. 530, Retaining Walls Example 3, Step 8, first column, insert text as shown below: Discussion Calculation ACI 318M-14 Step 8: Heel design Shear The base heel is designed for shear….not be linear as assumed (refer to Fig. E3.6) Note: In the 2019 edition of ACI 318 Section 13.3.6.3 was added, which sets the critical section for shear and flexure at the interface of the stem and footing. Checking the calculations: Stem: fVn = 226 kN > Vu = 92 kN Heel fVn = 258 kN < Vu = 227 kN OK

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OK

p. 549, Retaining Walls Example 4, Step 8, first column, insert text as shown below: ACI 318M-14 Discussion Calculation Step 8: Heel design Shear The base heel is designed for shear….not be linear as assumed (refer to Fig. E4.6) Note: In the 2019 edition of ACI 318 Section 13.3.6.3 was added, which sets the critical section for shear and flexure at the interface of the stem and footing. Checking the calculations: Stem: fVn = 170 kN > Vu = 62 kN Heel: fVn = 188 kN > Vu = 158 kN OK

p. 569, Retaining Walls Example 5, Step 8, first column, insert text as shown below: Discussion Calculation ACI 318M-14 Step 8: Heel design Shear The base heel is designed for shear…. Therefore, conservatively, it is not included in the calculation of shear strength. Note: In the 2019 edition of ACI 318 Section 13.3.6.3 was added, which sets the critical section for shear and flexure at the interface of the stem and footing. Checking the calculations: Stem: fVn = 226 kN > Vu = 90 kN Heel fVn = 223 kN > Vu = 209 kN OK

p. 584, Retaining Walls Example 6, Step 8, first column, insert text as shown below: ACI 318M-14 Discussion Calculation Step 8: Heel design Shear The base heel is designed for shear….not be linear as assumed (refer to Fig. E6.6) Note: In the 2019 edition of ACI 318 Section 13.3.6.3 was added, which sets the critical section for shear and flexure at the interface of the stem and footing. Checking the calculations: Stem: fVn = 136 kN > Vu = 31.5 kN Heel: fVn = 209 kN > Vu = 143 kN OK

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p. 664, Strut-and-tie Example 3, Step 11, second column, should be revised as: Discussion Calculation ACI 318M-14 Step 11: Node N2 Strut S2, a bottle-shaped strut with 23.9.1 reinforcement satisfying 23.5.323.5.1, is 54 mm wide at Node N3. At Node N2:

ACI 318M-14 Step 7: Pullout 17.3.3.c(ii)

p. 788, Anchorage Example 14, Step 7, third column, insert equation as shown below: Discussion Calculation For cast-in-headed studs with supplementary reinforcement, Condition A applies:

φ=0.75 φ Npn=(0.75)(1.0)(164.6 kN) = 123.4115.0 kN/stud φNpn = 123.4115.0 kN/stud > Nua,g = ~30 kN/stud OK

p. 789 Anchorage Example 14, Step 10, revise table as shown below: Discussion Calculation ACI 318M-14 Step 10: Tension forces summary ACI 318M Failure mode 17.4.1.2 Steel/stud 17.4.2.1 Concrete breakout/group 17.4.3.1 Concrete pullout/stud 17.4.4.1 Concrete side-face blowou t

Design Strength, kN φN sa 68

Ratio=Nua,g/φ φNn 0.44

Controls design? No

φNcbg

247

0.47

Yes

φNpn

123.4115.0

0.240.26

No

φNsa

NA

—

—

p. 792, Anchorage Example 14, Step 13, third column, should be revised as: ACI 318M-14 Discussion Calculation Step 13: Pryout 17.3.3c(i) Reduction factor φ for cast-in anchor φ=0.750.70 with supplementary reinforcement, φ Vcpg=(0.7570)(533 kN) = 399.7373 kN = 400 kN Condition A applies: φVcpg = 400 373 kN> Vua,g = 76 kN OK

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p. 793 Anchorage Example 14, Step 14, revise table as shown below: ACI 318M-14 Discussion Calculation Step 14: Shear force summary ACI 318M Failure mode 17.5.1.2 Steel/stud 17.5.2.1 Concrete breakout/group 17.5.3.1 Concrete pryout/group 17.4.4.1 Concrete splitting

Design Strength, kN 176.5 φV sa

φVn Ratio=Vua,g/φ 0.43

Controls design? No

φVcbg

159

0.48

Yes

φVcpg

400373

0.190.20

No

NA

—

—

—

p. 804, Anchorage Example 15, Step 13, third column, should be revised as: Discussion Calculation ACI 318M-14 Step 13: Pryout 17.3.3c(i) Reduction factor φ for adhesive anchor; f = 0. 75 Condition A fVcpg = (0 .75 )(255 kN )191 .2 kN =191 kN fVcpg =191 kN > Vua ,g = 38 kN

OK

f = 0. 7 fVcpg = (0 .70 )(255 kN )178 .5 kN = : 178 kN fVcpg =178 kN > Vua ,g = 38 kN

OK

p. 805 Anchorage Example 15, Step 14, revise table as shown below: Discussion Calculation ACI 318M-14 Step 14: Shear force summary ACI 318M Failure mode 17.5.1.2 Steel/stud 17.5.2.1 Concrete breakout/group 17.5.3.1 Concrete pryout/group 17.4.4.1 Concrete splitting

Design Strength, kN φV sa 19.5

Ratio=Vua,g/φ φVn 0.65

Controls design? Yes

φVcbg

156.2

0.24

No

φVcpg

191178

0.200.21

No

NA

—

—

—

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ACI 318M-14 Step 9: Pullout 17.3.3c(ii)

p. 832, Anchorage Example 18, Step 9, third column, should be revised as: Discussion Calculation f = 0. 75 fN pn = (0.70 )(1.0 )(437 kN ) = 328 kN

For cast-in headed bolt with supplementary reinforcement, Condition A applies:

fN pn > fN ua ,g N ua ,g = 516. 5 kN / 4 = 129. 1 kN OK

f = 0. 7 fN pn = (0.70 )(1.0 )(437 kN ) = 306 kN

Check that design strength is greater than required strength per anchor.

φNpn = 328 306 kN > φNua,g = 129.1 kN

OK

p. 836 Anchorage Example 18, Step 13, revise table as shown below: Discussion Calculation ACI 318M-14 Step 13: Tension force summary ACI 318M Failure mode 17.4.1 Steel strength/anchor 17.4.2 Concrete breakout with supplemental reinforcement/group 17.4.3 Concrete pullout/anchor

Design Strength, kN 270.7 φNsa

φNn Ratio=Nua,g/φ 0.48

Controls design? Yes

φNcbg

516.5

—

No—anchor reinforcement

φNpn

328306

0.390.48

No

p. 844, Anchorage Example 18, Step 18, third column, should be revised as: ACI 318M-14 Discussion Calculation Step 18: Concrete pryout 17.3.3c(ii) Reduction factor φ for cast-in anchor: φVcpg=(0.7570)(343.4 kN) = 257.7240 kN φ=0.75

p. 44 Anchorage Example 18, Step 19, revise table as shown below: ACI 318M-14 Discussion Calculation Step 19: Shear force summary Controls design? ACI 318M Failure mode Design Strength, kN φVn Ratio=Vua,g /φ 17.5.1 Steel strength/anchor 112.6 0.59 Yes φVsa 17.5.2 Concrete breakout Case 1 67 Supplemental with supplemental — φVcbg Case 2 133.7 reinforcement reinforcement/group 17.5.3 Concrete pryout/group 257.5240 0.520.56 Note 1 φVcpg Note 1: The length of the anchor embedment and the supplemental reinforcement would preclude concrete pryout failure mode.

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ERRATA as of January 7, 2020 p. 54, One-way Slab Example 2, Step 1, third column, equation revised as: ACI 318MDiscussion Calculation 14 Step 1: Geometry 7.3.1.1   6000 mm 

h≥

h≥ …

8

27

= 

27

 = 222 mm, say, 230 mm 

  6000 mm  =  = 215 mm, say, 230 mm 28  28 ...