CE 422
REINFORCED CONCRETE DESIGN

COURSE OUTLINE: I. Introduction to Reinforced Concrete II. Analysis and Design of Beams
Singly Reinforced Beams
Doubly Reinforced Beams
T-Beams
III. Shear and Diagonal Tension IV. Bonds, Development Length, Hooks and Splices Reinforcement V. Axially Loaded Columns VI. Eccentrically Loaded Columns (Short Columns Subjected to both Bending and Compression) VII. Long Columns VIII. Analysis and Design of Slabs
One Way Slab
Two Way Slab

References: Any Reinforced Concrete Book

Instructor: Engr. Emman Rey Quimson

INTRODUCTION

CONCRETE- a mixture of water, cement and gravel.
REINFORCED CONCRETE MATERIALS: 1. Cement 2. Aggregates- comprises 65-75 % of the mixture
Sand- aggregates passing the No. 4 sieve.
Gravel- aggregates larger than the sand. 3. Water- comprises 15-20% of the mixture. 4. Steel Reinforcement- deformed bar
-Grade 40 & Grade 60 are usual Grade of Reinforcement.
-Grade 40- 40 000 psi
-Grade 60- 60 000 psi TYPES OF STEEL REINFORCEMENT: 1. BILLET STEEL- newly made steel 2. RAIL STEEL-rerolled from old axle & rail.

STRENGTH OF CONCRETE PROPORTIONS: CLASS | STRENGTH | AA | 4000 psi | A | 3000 psi | B | 2500 psi | C | 2000 psi |

Unit weight of concrete-23.56 KN/m3 ; 2400 kg/ m3
Modulus of elasticity of concrete: @ 1500 kg/ m3 to 2500 kg/ m3 = Ec=Wc1.5 (0.043)f'c @normal weight (2400 kg/ m3)= Ec=4700 f'c

LOADING COMBINATION (Based on NSCP 1992) 1. DEADLOAD + LIVELOAD 1.4DL + 1.7LL 2. DEADLOAD + LIVELOAD + WIND LOAD 0.75(1.4DL + 1.7LL + 1.7W) 3. DEADLOAD + LIVELOAD + LATERAL PRESSURE 1.4DL + 1.7LL + 1.7H 4. DEADLOAD + LIVELOAD + EARTHQUAKE 0.75(1.4DL + 1.7LL + 1.9W)

DESIGN METHODS: 1. WSD- Working Strength Method 2. USD- Ultimate Strength Method

STRENGTH REDUCTION FACTORS, Ø

NATURE OF LOADING | Ø | FLEXURE/BENDING | 0.90 | SHEAR | 0.85 | COMPRESSION,SPIRALLY REINFORCED | 0.75 | COMPRESSION,TIED REINFORCEMENT | 0.70 | BEARING ON CONCRETE | 0.70 |

BASIC THEORIES, CONCEPTS AND PRINCIPLES: * If the beam is purely homogeneous and elastic, the flexural stress formula f=MC/I could be used in analysis, but since , at ultimate load, a beam is never homogeneous and elastic, the basic flexural stress formula could not be used. * Assumptions in the Design and Analysis of Flexural Members (Members Subject to Bending) 1. The tensile strength of the concrete is neglected and the tensile reinforcement is assumed to take all the tensile forces. 2. The stress in the reinforcement below specified yield strength, fy, is assumed to be equal to the modulus of elasticity of the steel, Es, times the steel strain Єs. 3. The concrete stress is assumed to be equal to 0.85f’c over the equivalent compression zone of the concrete. 4. The strain distribution is assumed to be linear. 5. The maximum usable strain in the concrete shall be equal to 0.003.

REINFORCED CONCRETE BEHAVIOR

Where: a depth of compression block c distance from the extreme concrete Compression fiber to the neutral axis N.A d depth of the beam b width of the beam As area of tension steel reinforcement = ρbd f’c specified compressive strength of concrete fy specified yield strength of steel fs calculated stress of steel below yield point Ey strain of steel at yield point = fy/Es Es strain of steel below yield point = fs/Es β factor stated in the National Standard Code of the Philippines ρ ratio of steel reinforcement= As/bd Es 200 000 MPa Slope; stress/strain fy/Es

NSCP 2001 for f’c > 30 MPa β1= 0.85-(0.05/7)(f’c-30) for f’c ≤ 30 MPa β1= 0.85

NSCP 1992 for f’c > 30 MPa β1= 0.85-0.008(f’c-30) for f’c ≤ 30 MPa β1= 0.85

NOMINAL MOMENT CAPACITY OF BEAM

From the figure above(stress diagram)
∑M@pt T=0
Mn=C(d-a2)
Mn=0.85f’c ab (d-a2) eq. 1
Solving for a:
∑Fh=0
C=T
0.85f’c ab= Asfy a=Asfy0.85f'cab a=Asfy0.85f'cab (dd) a=Asfy(d)0.85f'c(bd) since ρ=Asbd a=ρfyd0.85f'c Let ω=ρfydf'c a= ωd0.85 eq. 2

Substitute ❷ to ❶:
Mn=0.85f’c( ωd0.85)b (d-12( ωd0.85)) Mn=f’c bd2ω(1-.59ω) |

3 MODES OF FAILURE:

1. BALANCED DESIGN
A designed so proportional that the max stresses of concrete ( w/ strain of .003 ) & of steel ( w/ strain of fyEs ) are reached simultaneously as the ultimate load is reached ,causing them (concrete and steel) to fail simultaneously.

2. OVER- REINFORCED DESIGN
A design balanced in which the steel reinforcement is more than what is required for balanced condition.As load is increased, the deflection in the steel are not noticeable rete but the concrete is already overstressed. The steel under this condition will not yield at failure.

3. UNDER_ REINFORCED DESIGN
A design in which the steel reinforcement is less than what is required for balanced condition. If the ultimate load is approached, the steel will attain noticeable deflection while the concrete is still under- stressed. As load is further increased, steel will yield wl excessive deformations and deflections. Failure under this condition is ductile.

THE BALANCED STEEL RATIO

From the strain diagram: By ratio and proportion:

Cbd = 0.0030.003+ey
Cbd = 0.0030.003+FyEs
But Es = 200, 000 MPa

Cbd = 0.0030.003+.003200 000+fy200 000

SINGLY REINFORCED DESIGN

. DESIGN OF SINGLY REINFORCED BEAM Mu= Øf’cbd2 ω (1-0.59 ω) where: Mu= 1.4MDL + 1.7MLL
Ø= 0.90 ω= ρfyf'c where: ρ= ρassume=0.5ρmax where: ρmax=0.75 ρb ρb=
SAMPLE PROBLEM 1. Design a rectangular beam that will carry a factored moment Mu= 542.40KN-m, f’c= 27.58Mpa, fy= 275.80Mpa
Solution:

Mu= Øf’cbd ω 2(1-0.59 ω) Mu= 542.40KN-m f’c= 27.58MPa fy= 275.80MPa ω= ρfyf'c ρassume= 0.5ρmax , ρmax=0.75 ρb ρb= 0.85 f'c β1fy (600600+fy) Note: >DESIGNING (one layer)
The distance from the centroid of the tension bars to the extreme comp. fiber is 63.5mm. f’c < 30MPa , β1= 0.85 ρb= 0.85 (27.58)(0.85)275.80 (600600+275.80) ρb= 0.049 ρassume= (0.5)(0.75)(0.049) ρassume= 0.018 ω= ρfyf'c= 0.018(275.8027.58) ω= 0.18 Subst. values: b= d2 542.40 x 10002 = 0.90(27.58)bd2(0.18)(1-.59(0.18)) 542.40 x 10002 = 0.90(27.58)(0.5d)(d2)(0.18)(1-.59(0.18)) By calcu: d= 647.64mm = 650mm b= 0.5d = 0.5(650)= 325mm Rn= Mnbd2 Mn= Mu/Ø = 542.40/.90= 602.67KN-m Rn= 602.67(325)(650)2= 4.39MPa Revised ρ: ρ= 0.85 f'c fy [ 1- 1-2Rn0.85f'c ] >Clear spacing bet. bars; Is 25mm or the bar diameter, w/c ever is larger > Clear spacing bet. layers of bars should be 25mm ρ= 0.85 (27.58)275.80 [ 1- 1-24.3920.85(27.58) ] ρ= 0.018 Check: ρmin=1.4fy ; if ρ assume > ρ minimum ∴ use ρ assume 2. Design the most economical rectangular section of a beam using the strength method. The beam is simply supported with a span of 12m and it is to carry LL of 19 KN/m and DL of 11.67 KN/m (excluding self-wt.), f’c= 27.58 MPa, fy= 413.70 MPa.
Solution:
Mn= ∅f’cbd2 ω (1-0.59 ω) Mn= 1.4MDL + 1.7MLL