1

DATA SHEET FOR FINAL EXAM Reinforced Concrete Design II

A- R. C. Members subjected to Flexural Moments 1. Compression zone factor 𝛽1 = 0.85 𝑓𝑜𝑟 𝑓𝑐

′ ≤ 28 𝑀𝑃𝑎

= 0.85 − 0.05 (𝑓𝑐

′−28

7) 𝑓𝑜𝑟 28 < 𝑓𝑐

′ ≤ 56 𝑀𝑃𝑎

= 0.65 𝑓𝑜𝑟 𝑓𝑐′ > 56 𝑀𝑃𝑎

2. Reinforcement ratio limitation, singly reinforced sections (rectangular and T section “tension in the web”)

𝜌𝑚𝑖𝑛 = 1

4

√𝑓𝑐′

𝑓𝑦≥

1.4

𝑓𝑦, 𝐴𝑠,𝑚𝑖𝑛 = 𝑏𝑑 𝜌𝑚𝑎𝑥 = 0.319𝛽1

𝑓𝑐′

𝑓𝑦

3. reinforcement ratio limitation, doubled reinforced sections (rectangular and T section “tension in the web”)

𝜌𝑚𝑖𝑛 = 1

4

√𝑓𝑐′

𝑓𝑦≥

1.4

𝑓𝑦, 𝐴𝑠,𝑚𝑖𝑛 = 𝑏𝑑 𝜌𝑚𝑎𝑥 = 0.04

4. Reinforcement ratio limitation, singly reinforced T section “tension in the flange”

𝜌𝑚𝑖𝑛 = 1

4

√𝑓𝑐′

𝑓𝑦≥

1.4

𝑓𝑦, 𝐴𝑠,𝑚𝑖𝑛 = 𝜌𝑚𝑖𝑛(2𝑏𝑤 𝑜𝑟 𝑏𝑒)𝑑 𝜌𝑚𝑎𝑥 = 0.319𝛽1

𝑓𝑐′

𝑓𝑦

The least of 2𝑏𝑤 𝑜𝑟 𝑏𝑒

Where 𝑏𝑒

5. Material Reduction Factor

φ = 0.9 for tension controlled sections

φ = 0.65 for compression controlled sections

𝜑 = 0.65 + 0.25 ∗{𝜖𝑠−𝜖𝑦}

{𝜖𝑡−𝜖𝑦}

6. Data used for designing a section with tension steel only subjected to flexure moment 𝑀𝑢:

Reinforcing Index 𝜔 = 𝜌 𝑓𝑦

𝑓𝑐′ with 𝜔 =

𝛽1

4

Flexural Resistance Factor 𝑅 = 𝜔 𝑓𝑐′(1 − 0.59𝜔)

Nominal Moment capacity 𝑀𝑛 = 𝑅𝑏𝑑2

Ultimate Moment capacity 𝑀𝑢 = 𝜑𝑅𝑏𝑑2, φ = 0.9

7. Data used for designing a section with doubled steel (tension and compression reinforcements) subjected to flexure

moment 𝑀𝑢:when 𝜔 > 0.319𝛽1

Single Moment capacity(𝑀𝑢 ) 𝑠 = 𝜑𝑅𝑏𝑑2, φ = 0.9 and 𝑅 = 𝜔 𝑓𝑐′(1 − 0.59𝜔), 𝑤𝑖𝑡ℎ 𝜔 = 0.319𝛽1

Remaining moment to be resist by compression steel and part of tension reinforcements (𝐴𝑠) 1

𝑀𝑟 = 𝑀𝑢 − (𝑀𝑢 ) 𝑠 = 𝜑𝐴𝑠′ [𝑓𝑦 𝑜𝑟 𝑓𝑠

′ − 0.85𝑓𝑐′](𝑑 − 𝑑′)

2

B- Shear Strength of Concrete of Beams

𝜑𝑉𝑐 = 𝜑𝜆√𝑓𝑐

′

6𝑏𝑑 With 𝜆 = 1 (N. W. C), 𝜆 = 0.85 (S. L.W. C), 𝜆 = 0.75 (A. L.W. C)

(𝑉𝑢)𝑟 = (𝑉𝑢)𝑐 ∗

𝐿2

− (𝑑 +𝑐2

)

𝐿2

𝑉𝑢 = 𝜑𝑉𝒄 + 𝜑𝑉𝒔

𝜑𝑉𝑠 = 𝜑𝐴𝑣𝑓𝑦𝑡(𝑠𝑖𝑛𝛼 + 𝑐𝑜𝑠𝛼)𝑑

𝑆

Minimum shear reinf. 𝐴𝑣 𝑚𝑖𝑛 =1

16

√𝑓𝑐′

𝑓𝑦𝑡𝑏𝑆 𝑂𝑅 0.35

𝑏𝑆

𝑓𝑦𝑡

Case 1 If 𝑉𝑢 ≤ 0.5 ∅𝑉𝒄 ---> No reinforcements needed

Case 2 If 0.5 ∅𝑉𝒄 ≤ V𝑢 ≤ ∅𝑉𝒄 ---> Minimum reinforcements

Case 3 If 𝑉𝑢 ≤ ∅ (𝑉𝑐 +√𝑓𝑐

′

3𝑏𝑑) , 𝑆𝑚𝑎𝑥 =

𝑑

2 and ≤ 600 𝑚𝑚

Case 4 If 𝑉𝑢 > ∅ (𝑉𝑐 +√𝑓𝑐

′

3𝑏𝑑) , 𝑆𝑚𝑎𝑥 =

𝑑

4 and ≤ 300 𝑚𝑚

Case 5 If 𝑉𝑢 > ∅ (𝑉𝑐 +2

3√𝑓𝑐

′𝑏𝑑) , 𝑆𝑒𝑐𝑡𝑖𝑜𝑛 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑠 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑑

Table B.5 Minimum Bar Width(mm) for Beams

Size

of Bars

Number of Bars in Single Layer of reinforcement

2 3 4 5 6 7 8

#13 175 213 251 288 326 364 401

#16 178 219 260 301 342 383 424

#19 182 226 270 314 358 402 446

#22 185 232 279 326 373 421 468

#25 188 239 290 341 391 442 493

#29 195 252 310 367 424 482 539

#32 202 267 331 396 460 526 590

#36 209 281 353 424 496 567 639

#43 228 314 400 486 572 658 744

#57 271 386 501 615 730 844 959

C- ACI Moment and Shear Coefficients (used for indeterminate continuous beams or one way

slabs)

Bar

Designation

Nominal

dia.

(mm)

Area

(mm2)

#10 9.5

71

#13 12.7

129

#16 15.9

199

#19 19.1

284

#22 22.2

387

#25 25.4

510

#29 28.7

645

#32 32.3

819

#36 35.8

1006

#43 43.0

1452

#57 57.3

2581

3

D- Minimum thickness of beams and one way slabs using Empirical Method

Cantilevers

4

ACI code provisions to control crack width

The ACI code provision on crack control limits the spacing between tension reinforcement to values

given by :

S max = 380∗280

𝑓𝑠− 2.5𝐶𝑐 ≯

300∗280

𝑓𝑠≯ 450𝑚𝑚

Where : 𝑓𝑠 = 2/3𝑓𝑦

Cc = Clear cover of the steel reinforcement (mm)

S = spacing of tension reinforcement closest to the tension face (mm)

Minimum reinforcement area to control the crack: Ash=0.0018bh for fy=420Mpa

E – Two Way Slabs 1. Beam –slab stiffness ratio

2. Beam cross Sections

3. Beam stiffness

scs

bcb

scs

bcbf

IE

IE

/lIE

/lIE

4

4

5

4. Slab thickness

(a)

β = Ratio of the long to short span of the panel Note: At discontinuous edge, beams shall have minimum stiffness ratio:

Otherwise, slab thickness as determined by equations must be increased by 10%

4.Paneled beams

Where r is the ratio of long to short side of edge beams.

WB=(Wu)(S)(sin)

5. Statical Moment Mo

.

6. Moments in the Equivalent Frame

For interior panels, the total factored moment Mo is divided into:

- Positive moment = 0.35 Mo

- Negative moment = 0.65 Mo

For exterior panels, the total factored moment Mo is divided into M+ and M- according to table 13.6.3.3

2m

(b) Slabs with beams between interior supports

22.0 m

8

2

n12u0

llqM X

8

2

n21u

0

llqM Y

cn

n

2

u

0.886d h using calc. columns,circular for

columnsbetween span clear

strip theof width e transvers

areaunit per load factored

l

l

l

q

6

7. Distribution of Moments between column and middle strips

Where: x is the short side of the rectangle

y is the long side of the rectangle

The above Tables 4-3, 4-4 and 4-5 can also be calculated from the following relations:

(i) % Column strip M-v

factors for interior spans

= 75 + 30 (𝛼1𝐿2

𝐿1

) (1 −𝐿2

𝐿1

)

(ii) % Column strip M-v

factors for exterior spans

= 100 − 10𝛽𝑡 + 12𝛽𝑡 (𝛼1𝐿2

𝐿1

) (1 −𝐿2

𝐿1

)

(iii) % Column strip M+v

factors for interior and exterior spans

= 60 + 30 (𝛼1𝐿2

𝐿1

) (1.5 −𝐿2

𝐿1

)

8. Moment Transfer to columns For interior columns

qDu

and qLu

= the factored DL and LL on the longer span.

q’Du

= The factored DL on the shorter span adjacent to the column.

Punching shear equations (least of the followings):

3

63.01

2

3

scs

cbt

scs

bcb1

yx

y

xC

IE

CE

IE

IE

7

1. Punching Shear strength by stirrups

2. Punching Shear strength by shear studs

3. Moment transfer

4.

8

5. Polar moment of inertia Jc

6.

7. Equivalent Stiffness to be used for EFM