Lecture (1) Introduction to Material properties And Analysis of Reinforced Concrete Sections By Dr. Islam M. El‐Habbal
Lecture (1) Introduction to Material properties
And Analysis of Reinforced Concrete Sections
By
Dr. Islam M. El‐Habbal
Stress‐Strain Curve of Concrete
0 0.001 0.002 0.003 0.004
Stre
ss
Strain
fcu Softening
0 0.001 0.002 0.003 0.004St
ress
Strain
fcu
Crushing Strain
Design Stress‐strain Diagram
0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035
Stre
ss
Strain
Tangent Modulus or Young’s Modulus
0.30 fcu
fcu
fcu/γc
Factor of Safety =1.50
Stress
Strain
fy
fu
Stress‐Strain Curve of Steel
εy εu
Design Stress‐Strain Curve of Steel Stress
Strain
fy/γs
fy
εy/γs εy
Factor of Safety =1.15
b
d t
As
Types of RC sections 1. Rectangular Section
Is used when slab is under Tension
d t
b
As
ts
B
2. T‐ Section
Is used when slab is under compression
1. Analysis of Rectangular Sections
2. Analysis of T‐section.
Analysis of Rectangular Singly Reinforced Sections using 1st principles
From Equilibrium:
Cu = Tu ……………………. (1)
0.67 fcu/γc *a*b = As * fy/γs get (a) check a/d≥ 0.10
c = a/0.80
Mu = Cu * yct or Mu = Tu * yct …………… (2)
Mu = 0.67 fcu/γc *a*b *(d‐a/2) or Mu = As * fy/γs *(d‐a/2) Get (Mu )
b
d t Mu
εy/γs
0.003
Strain Diagram
N.A. c
0.67 fcu/γc
Stress Diagram
0.67 fcu/γc a Cu
Tu
Idialized Stress Diagram
As
yct
Analysis of T‐Sections Singly Reinforced using 1st principles
d t
b
Mu
εy/γs
0.003
Strain Diagram
N.A. c
0.67 fcu/γc
Stress Diagram
0.67 fcu/γc a Cu
Tu
Idialized Stress Diagram
As
yct
ts
B
From Equilibrium:
Cu = Tu ……………………. (1)
0.67 fcu/γc *a*B = As * fy/γs get (a) if a> ts let a=ts check a/d ≥ 0.10 c = a/0.80
Mu = Cu * yct or Mu = Tu * yct …………… (2)
Mu = 0.67 fcu/γc *a*B *(d‐a/2) or Mu = As * fy/γs *(d‐a/2) Get (Mu )
1. Analysis of Rectangular Sections
Analysis of Rectangular Doubly Reinforced Sections using 1st principles
From Equilibrium:
Cu +Cs\= Tu
0.67 fcu/γc *a*b +As\ * fs\= As * fs ……………………. (1)
Assume that As & As\ are @ yield
0.67 fcu/γc *a*b +As\ * fy/γs = As * fy/γs get (a) check a/d≥ 0.10
c = a/0.80
b
Mu
εy/γs
0.003
Strain Diagram
N.A. c
0.67 fcu/γc
Stress Diagram
0.67 fcu/γc a Cu
Tu
Idialized Stress Diagram
As
yct
As\
d t
d’ εs\
Cs\
b
Mu
εs
0.003
Strain Diagram
N.A. c
0.67 fcu/γc
Stress Diagram
0.67 fcu/γc a Cu
Tu
Idialized Stress Diagram
As
yct
As\
d t
d’ εs\
Cs\
Check on Assumption: Since E=f/ε where E=Young’s modulus & f=stress & ε=strain So E=fy/εy where fy = yield stress & εy = yield strain & E=2000 t/cm2 = 2 x 105 MPa = 2 x 105 N/mm2 Hence get εy From symmetry of triangles: 0.003/c = εs
\ / (c‐d’) εs
\ = 0.003 * (c‐d’) / c
0.003/c = εs
/ (d‐c) εs = 0.003 * (d‐c) / c
Check of Assumption
εs\ ≥ εy/γs
εs ≥ εy/γs
εs\ < εy/γs
εs ≥ εy/γs εs
\ ≥ εy/γs
εs < εy/γs εs
\ < εy/γs
εs < εy/γs
No change in Calculations. Get Mu directly
Replace As’ fy By As’ fs ’ . get fs ’(c). Solve fs ’(c) & (1) to get C. Solve fs ’(c) and Mu(c) sim.
Replace As fy By As fs . get fs(c). Solve fs (c) & (1) to get C. Solve fs(c) and Mu(c) sim.
Replace As’ fy & As fy By As’ fs ’ & As fs . get fs ’(c), fs(c). Solve fs ’(c), fs (c) & (1) to get C. Solve fs ’(c), fs(c) and Mu(c) sim.
Get (Mu )
If εs\ ≥ εy/γs & εs ≥ εy/γs:
Mu = Cu * yct + Cs\ * (d‐d’) or Mu = Tu * yct + Cs
\ * (a/2‐d’) …………… (2)
Mu = 0.67 fcu/γc *a*b *(d‐a/2) + As\ * fy/γs * (d‐d’)
or Mu = As * fy/γs *(d‐a/2) + As\ * fy/γs * (a/2‐d’)
b
Mu
εs
0.003
Strain Diagram
N.A. c
0.67 fcu/γc
Stress Diagram
0.67 fcu/γc a Cu
Tu
Idialized Stress Diagram
As
yct
As\
d t
d’ εs\
Cs\
Check of Assumption
εs\ ≥ εy/γs
εs ≥ εy/γs
εs\ < εy/γs
εs ≥ εy/γs εs
\ ≥ εy/γs
εs < εy/γs εs
\ < εy/γs
εs < εy/γs
If εs\ < εy/γs & εs ≥ εy/γs:
fs\ = εs\ * E fs\ = 0.003 * E * (c‐d’) / c ………….. (2)
Solve (1) & (2) together to Get C C=a/0.8 check a/d ≥ 0.10 get fs\ Mu = Cu * yct + Cs
\ * (d‐d’) or Mu = Tu * yct + Cs
\ * (a/2‐d’) …………… (3)
Mu = 0.67 fcu/γc *a*b *(d‐a/2) + As\ * fs\ * (d‐d’)
or Mu = As * fy/γs *(d‐a/2) + As\ * fs\ * (a/2‐d’)
Solve (2) & (3) together and Get (Mu )
b
Mu
εs
0.003
Strain Diagram
N.A. c
0.67 fcu/γc
Stress Diagram
0.67 fcu/γc a Cu
Tu
Idialized Stress Diagram
As
yct
As\
d t
d’ εs\
Cs\
Check of Assumption
εs\ ≥ εy/γs
εs ≥ εy/γs
εs\ < εy/γs
εs ≥ εy/γs εs
\ ≥ εy/γs
εs < εy/γs εs
\ < εy/γs
εs < εy/γs
If εs\ ≥ εy/γs & εs < εy/γs:
fs = εs * E fs = 0.003 * E * (d‐c) / c ………….. (2)
Solve (1) & (2) together and Get C C=a/0.8 check a/d ≥ 0.10 get fs Mu = Cu * yct + Cs
\ * (d‐d’) or Mu = Tu * yct + Cs
\ * (a/2‐d’) …………… (3)
Mu = 0.67 fcu/γc *a*b *(d‐a/2) + As\ * fy/γs * (d‐d’)
or Mu = As * fs *(d‐a/2) + As\ * fy/γs * (a/2‐d’)
Solve (2) & (3) together and Get (Mu )
b
Mu
εs
0.003
Strain Diagram
N.A. c
0.67 fcu/γc
Stress Diagram
0.67 fcu/γc a Cu
Tu
Idialized Stress Diagram
As
yct
As\
d t
d’ εs\
Cs\
Check of Assumption
εs\ ≥ εy/γs
εs ≥ εy/γs
εs\ < εy/γs
εs ≥ εy/γs εs
\ ≥ εy/γs
εs < εy/γs εs
\ < εy/γs
εs < εy/γs
If εs\ < εy/γs & εs < εy/γs:
fs = εs * E & fs\ = εs\ * E fs = 0.003 * E * (d‐c) / c ………….. (2)
fs\ = 0.003 * E * (c‐d’) / c ………….. (3)
Solve (1) , (2) & (3) together and Get C C=a/0.8 check a/d ≥ 0.10 get fs & fs\ Mu = Cu * yct + Cs
\ * (d‐d’) or Mu = Tu * yct + Cs
\ * (a/2‐d’) …………… (4)
Mu = 0.67 fcu/γc *a*b *(d‐a/2) + As\ * fs\ * (d‐d’)
or Mu = As * fs *(d‐a/2) + As\ * fs\ * (a/2‐d’)
Solve (2) , (3) & (4) together and Get (Mu )
b
Mu
εs
0.003
Strain Diagram
N.A. c
0.67 fcu/γc
Stress Diagram
0.67 fcu/γc a Cu
Tu
Idialized Stress Diagram
As
yct
As\
d t
d’ εs\
Cs\
Check of Assumption
εs\ ≥ εy/γs
εs ≥ εy/γs
εs\ < εy/γs
εs ≥ εy/γs εs
\ ≥ εy/γs
εs < εy/γs εs
\ < εy/γs
εs < εy/γs