Engr 327 reinforced concrete design i dr solomon

ENGR 327 Reinforced Concrete Design I
Dr. Solomon Tesfamariam
4-13
Example 2: Find M
r
using first principles.
Given Æ
µ
′
= 30 MPa
f
y
= 400 MPa
Concrete cover = 40 mm
d = 600 – 40 – 11.3 – 25.2/2
As = 4 - 25M
600 mm
500 mm
40 mm
10M Stirrup

ENGR 327 Reinforced Concrete Design I
Dr. Solomon Tesfamariam
4-14
4.5 A
NALYSIS FOR
O
VER
-R
EINFORCED
F
AILURE
° For over-reinforced conditions: ε
c
= ε
cu
ε
s
< ε
y
From strain compatibility:
´
»
´
µ¶
¸
¼ ¾ ¹
¹
Thus,
´
»
¸ ½
¼ ¾ ¹
¹
¿ ´
µ¶
´
»
¸ ½
¼ ¾ ¹
¹
¿ ´
µ¶
Æ
»
¸ D
»
½
Ç
Å
¼
0
¾ 1¿ ´
µ¶
Compression: 1
²
¸ Ä
Å
·
µ
Æ
µ
2
03
Tension: 4
²
¸ ·
»
È
»
Æ
»
¸ ·
»
È
»
D
»
F
@
=
!
?
¾ 1G ´
µ¶

ENGR 327 Reinforced Concrete Design I
Dr. Solomon Tesfamariam
4-15
Equilibrium: 1
²
¸ 4
²
Ä
Å
·
µ
Æ
µ
2
03 ¸ ·
»
È
»
D
»
F
@
=
!
?
¾ 1G ´
µ¶
Rearranging:
Ä
Å
·
µ
Æ
µ
2
03 ¸
·
»
È
»
D
»
´
µ¶
Ç
Å
¼
0
¾ ·
»
È
»
D
»
´
µ¶
ÀÄ
Å
·
µ
Æ
µ
2
3Â0
H
º À·
»
È
»
D
»
´
µ¶
Â0 ¾ À·
»
È
»
D
»
´
µ¶
Ç
Å
¼Â ¸ 0
Solve quadratic for “a”
¹ ¸
?
@
=
to check for ε
s
< ε
y
Check:
µ
!
:
/’’
/’’%³
&
Lever arm: z = d - a/2
Moment
Resistance: Mr = 4
²
A
Mr = ·
»
È
»
Æ
»
À¼ ¾ 0/2Â
or
Mr = 1
²
A
Mr = Ä
Å
·
µ
Æ
µ
2
03À¼ ¾ 0/2Â

ENGR 327 Reinforced Concrete Design I
Dr. Solomon Tesfamariam
4-16
Example 3: Find M
r
using first principles.
Given Æ
µ
′
= 30 MPa
f
y
= 400 MPa
Concrete cover = 40 mm
d = 600 – 40 – 11.3 – 35.7 – 50/2
As = 12 - 35M
600 mm
500 mm
40 mm
10M Stirrup
50 mm

ENGR 327 Reinforced Concrete Design I
Dr. Solomon Tesfamariam
4-17
4.5 C
RACKING
M
OMENT
& M
INIMUM
R
EINFORCEMENT
±
µ²
¸
Æ
²
I
J
Ã
K
¸ Æ
²
L
L ¸ Section modulus ¸
3Y
H
6
Æ
²
¸ 0.6·
µ
[\Æ
µ
′
° Independent of the steel reinforcement area
È
]^_‘
= minimum amount of steel to avoid sudden failure when concrete cracks
To avoid brittle failure at first cracking:
±
²
: 1.2±
µ²
Clause 10.5.1.1
Ê
^_‘
¸
’.Ha³
#
>
³
&
Clause 10.5.1.2
or
È
]^_‘
¸
’.Ha³
#
>
³
&
3
K
h
where,
3
K
= width of tension zone of the section

ENGR 327 Reinforced Concrete Design I
Dr. Solomon Tesfamariam
4-18
4.6 H
ANDBOOK
A
NALYSIS
/ D
ESIGN
T
ABLES
1 ¸ Ä
Å
·
µ
Æ
µ
2
03
4 ¸ ·
»
Æ
Á
Ê3¼
0 ¸
·
»
Æ
Á
Ê
Ä
Å
·
µ
Æ
µ
2
¼
M
r
= 4À¼ ¾ 0/2Â
¸ ·
»
Æ
Á
Ê3¼ 5¼ ¾
·
»
Æ
Á
Ê3
2Ä
Å
·
µ
Æ
µ
2
¼8
¸ Ê·
»
Æ
Á
51 ¾
·
»
Æ
Á
Ê
2Ä
Å
·
µ
Æ
µ
2
8 3¼
H
Let,
b
²
¸ Ê·
»
Æ
Á
51 ¾
·
»
Æ
Á
Ê
2Ä
Å
·
µ
Æ
µ
2
8
±
²
¸ b
²
3¼
H
Values of K
r
are tabulated in Table 2.1 of the CAC Handbook
Function of f
y
, Æ
µ
′
, Ê
Ê ¸
È
»
3¼

ENGR 327 Reinforced Concrete Design I
Dr. Solomon Tesfamariam
4-19