Ultimate Strength Design (Doubly)

Reinforced Concrete Design ( RCD – 1 )Monday 5:00 – 9:00 / Thursday 6:00 – 9:00Engr. Alberto S. Cañete - Professor

Ultimate Stress Design (USD)Lecture Notes

2k6 – 2k7

DOUBLY REINFORCED RECTANGULAR BEAMS

ANALYSIS:



2k6 – 2k7

ASIDE

fsc = (c – d’) * 600 / c > fy (Step 7)< fy (Step 8)



2k6 – 2k7

DESIGN :

SAMPLE PROBLEM:

ASIDE



2k6 – 2k7



2k6 – 2k7

Determine the maximum WL that the beam can carry if WDL = 10 kN/m. f’c =28Mpa using Grade 60 bars.

Solution:

BEAMS WITH IRREGULAR CROSS-SECTIONS

T-BEAMS

Effective Width, be (NSCP Art. 408.11, p.4-26)

A. Interior Beams1. L/42. bw + 16ts3. c-c spacing

B. Exterior Beams1. bw + L/122. bw + 6ts3. (c-c spacing + bw) 2

IRREGULAR BEAMS



2k6 – 2k7



2k6 – 2k7

ANALYSIS:

Given: be, bw, d, ts, As, f”c, fyFind: Mu = ø MnSolution:

1. β1, ρmax

2. Assume a = ts

Cc=0.85f”c ts be

Ts=Asfy 3. If Ts < Cc => a < ts => singly Ts > Cc => a > ts => irregular

If irregular

4.)

5.) Asw = As - Asf

6.) As = Asf + Asw Mn = Mf + Mw

7.)

8.) If ρw < ρmax => ok! No reduction in Asw

If ρw > ρmax => Asw = ρmaxbwd

9.)

10. Mw =

11. Mu = ø (Mf + Mw)

DESIGN:

Given : be, bw, ts, d, f’c, fy, MuFind: As

1.) β1, ρmax

2.) Assume a = ts

3.) If Mu < ø Mn a < ts singly If Mu > ø Mn a > ts irregular

If irregular

4.)



2k6 – 2k7

5.)

6.) cffym

'85.0

7.)

8.)

9.) If ρw > ρmax increase beam size bw,d ρw < ρmax ok! Req’d Asw = ρwbwd 10. As = Asf + Asw

SAMPLE PROBLEM:

Design an interior beam of a floor system having a simple span of 18m, slab thickness of 100 mm, c-c spacing of 2.5 m, stem width of 300 mm, d= 630 mm, f’c = 20 MPa, h= 700mm and Grade 40 bars.

The floor system is to carry super imposed loads listed below.

Dead Loads:1. Ceiling = 0.5 kPa2. Floor Finish = 1.8 kPa3. Movable Partition = 1.0 kPa

Live Loads = 4.8 kPa

Solution:

Dead Loads:Ceiling = 0.5 kPa (2.5 m) = 1.25 kN/mFloor fin = 1.8 kPa (2.5m) = 4.5 kN/mMovable Par= 1.0 kPa (2.5m) = 2.5 kN/mSlab = 0.1 (2.5) (24) = 6 kN/mBeam web = 24 (0.3) (0.7-0.1) = 4.32 kN/m

ΣDL = 18.57 kN/m

Live Load = 4.8 ( 2.5) = 12 kN/m :(2.5) is tributary widthWu = 1.4 (18.57) + 1.7 (12)Wu = 46.398 kN/m



2k6 – 2k7

Effective Width

a.) mLbe 5.44

184

b.) be = bw + 16ts = 300mm + 16(100mm) = 1.9mc.) be = 2.5 m.Therefore.. be = 1.9m (smallest value governs)

1.) β1 = 0.85

ρmax = 0.026895

2.) 2'85.0 tsatsbecfMn

= 0.9 (0.85) (20) (1900) (100) (630-100/2) x 10-6

= 1686.06 kNm (compare to Mu) 3.)

4.)

5.) kNmMw 31.5106.15779.012.1879

6.) 235.152085.0

276'85.0

cf

fym

7.)

8.)

9.) ρw < ρmax ok! Req’d Asw = ρwbwd = 0.018224 (300)(630) = 3444.34 mm2

10.)

3-25 mm

Wu

0.4 Wu0.4 Wu



2k6 – 2k7

QUIZ SINGLY AND DOUBLY

At B:

3-25 mm

6-25 mm

3-25 mm



2k6 – 2k7

At C:

6-25 mm



2k6 – 2k7

QUIZ 5: DOUBLY REINFORCED RECTANGULAR BEAM & IRREGULAR SECTIONS

1.) Determine the maximum safe live load WL that the beam below can carry. Use f’c = 32 MPa and Grade 60 bars. Assume that the total dead load WD = 25 kN/m.



2k6 – 2k7Reinforced Concrete Design ( RCD – 1 )Monday 5:00 – 9:00 / Thursday 6:00 – 9:00Engr. Alberto S. Cañete - Professor


2k6 – 2k7

2.) Determine the ultimate moment capacities of the sections shown below. Use f’c = 23 MPa and Grade 40 bars.



2k6 – 2k7



2k6 – 2k7

Ultimate Strength Design (Doubly)

Documents