Post on 01-May-2017
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:
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
ASIDE
fsc = (c – d’) * 600 / c > fy (Step 7)< fy (Step 8)
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
DESIGN :
SAMPLE PROBLEM:
ASIDE
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
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
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
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
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
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.)
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
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
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
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
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
QUIZ SINGLY AND DOUBLY
At B:
3-25 mm
6-25 mm
3-25 mm
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
At C:
6-25 mm
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
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.
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 – 2k7Reinforced 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
2.) Determine the ultimate moment capacities of the sections shown below. Use f’c = 23 MPa and Grade 40 bars.
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
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