Tutorial 8 (Se e Example 6.

1 ) page 1/6

Re f. Calculations Output

SPECIFICATION

w kN/m
h

L
b

Effective span, L = 8.0 m

Characteristic Actions:
Permanent, g k = 25 kN/m (Excluding selfweight)
Variable, q k = 15 kN/m
Design life = 50 Years (Table 2.1 EN 1990)
Fire resistance = 1.5 hours (Sec. 5 EN 1992-1-2)
Table 4.1 Exposure classes = XC3
Materials :
3.1.2(3) Characteristic strength of concrete, f ck = = N/mm2 (Table F.1 EN 206)
3.2.2(2) Characteristic strength of steel, f yk = N/mm2
Characteristic strength of link, f yk = N/mm2
Unit weight of reinforced concrete = kN/m3 (Table A.1 EN 1991-1)
Assumed: f bar 1 = mm
f bar 2 = mm
f link = mm

SIZE
Overall depth, h = mm Use : b xh
Width, b = mm

DURABILITY, FIRE & BOND REQUIREMENTS

Table 4.2 Min. conc. cover regard to bond, C min,b = mm
Table 4.4N Min. conc. cover regard to durability, C min,dur = mm
4.4.1.2 Min. required axis distance for fire resistance, Table 5.5 EN 1992-1-2
a sd = + 10 =
Min. concrete cover regard to fire,
C min = a sd - f link - f bar/2
=
4.4.1.3 Allowance in design for deviation, D C dev =

4.4.1.1(2) Nominal cover, Use:

C nom = C min + DCdev = C nom

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Tutorial 8 (Se e Example 6.1 ) page 2/6

Re f. Calculations Output

LOADING & ANALYSIS

Beam selfweight = kN/m
Permanent load (Excluding selfweight) = kN/m
Characteristic permanent action, g k = kN/m
Characteristic variable action, q k = kN/m
Design action,
Table A1.2B w d = 1.35g k + 1.5q k =
: EN 1990 =
wd = kN/m

L = m

Shear Force,
V = w d L /2
=
Bending Moment,
M = w d L 2/8
M =

6.1 MAIN REINFORCEMENT

Effective depth, d'

d = h - C nom - f link - f bar = d

h
d' = Cnom + f link - f bar/2 =

Design bending moment, MEd = b

2
K = M / bd f ck
=
=
Redistribution = 0% Redistribution ratio, d = 1.0 Using : EC2
K bal = 0.454(d - k 1)/k 2 - 0.182[(d - k 1)/k2]2 k 1 = 0.44
= 0.363 (d - k 1) - 0.116 (d - k 1)2 k 2 = 1.25
= 0.167
K K bal

z = d [ 0.5 + 0.25 - K bal/1.134) ]

=
x = (d - z ) / 0.4 =
d '/x =

f sc < 0.87f yk

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Tutorial 8 (Se e Example 6.1 ) page 3/6

Re f. Calculations Output

Area of compression steel

A s' = (K - K bal) f ckbd 2 / 0.87f yk(d - d ')
=

= Use :

Area of tension steel

As = K bal f ckbd 2 / 0.87f yk z bal + A s'
=

=
Use :
9.2.1.1 Minimum and maximum reinforcement area,
A s,min = 0.26(f ctm/f yk) bd
=
=
=
As,max = 0.04Ac = 0.04 bh
=

6.2 SHEAR REINFORCEMENT

Design shear force, V Ed = kN

6.2.3 Concrete strut capacity

V Rd, max = 0.36b wdf ck(1 - f ck/250) / (cot q + tan q )
=
(cot q + tan q )
= kN q = 22 deg cot q = 2.5
= kN q = 45 deg cot q = 1.0

V Ed V Rd, max cot q = 2.5

V Ed V Rd, max cot q = 1.0
Therefore angle q 22o

= 0.5sin -1 [V Ed / 0.18b wdf ck(1- f ck/250)]

= 0.5sin -1

= 0.5sin -1

Use : = tan q cot q

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Tutorial 8 (Se e Example 6.1 ) page 4/6

Re f. Calculations Output

Shear links
A sw / s = V Ed / 0.78f ykd cot q
=
=
Try link : A sw =
Spacing, s =
=
Use :
9.2.2 (6) Max. spacing, s max = 0.75d = 0.75

9.2.2(5) Minimum links

A sw / s = 0.08f ck1/2b w / f yk
=
=
Try link : A sw =
Spacing, s = Use :
= 0.75d =

Shear resistance of minimum links

Vmin = (A sw/s )(0.78df yk cot q )
=
=

Links arrange me nt

x=
=

6.2.3(7) Additional longitudinal re inforce me nt

Additional tensile force,
DF td = 0.5V Ed cot q
=
M E d,max /z =
= DF td

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Tutorial 8 (Se e Example 6.1 ) page 5/6

Re f. Calculations Output

Additional longitudinal reinforcement, To be added to the

As = DF td / 0.87f yk main bar at support
=
=

7.4 DEFLECTION
Percentage of required tension reinforcement,
r = A s,req / bd
=
Reference reinforcement ratio,
r o = (f ck) 1/2 x 10-3
Percentage of required compression reinforcement,
r ' = A s ' ,req / bd
=
Table 7.4N Factor for structural system, K =
r ro Use equation

l r r
3/ 2

K 11 1.5 fck o 3.2 fck o 1 (1)

d r r

l ro 1 r'
K 11 1.5 fck fck (2)
d r r ' 12 r

Therefore basic span-effective depth ratio, l /d =

Modification factor for span,
=
Modification factor for steel area provided,
= A s,prov/A s,req =
Therefore allowable span-effective depth ratio,
(l /d )allowble =
Actual span-effective depth
(l /d )actual = (l /d )allowble

7.3 CRACKING
Table 7.1N Limiting crack width, w max = mm
Steel stress,
f yk Gk 0.3Qk 1
fs x
1.15 (1.35Gk 1.5Qk ) d

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Tutorial 8 (Se e Example 6.1 ) page 6/6

Re f. Calculations Output

=
=
Table 7.3N Max. allowable bar spacing =
Bar spacing,
s =
=

DETAILING

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Tutorial 9 (Se e Example 7.4 : Two-way Re straine d Slab) page 1/6

Re f. Calculations Output

SPECIFICATION

A B C D
7000 7000 7000

4000

4000

Plan vie w

Characteristic Actions:
Permanent, g k = 1.0 kN/m2 (Excluding selfweight)
Variable, q k = 3.0 kN/m2
Design life = 50 Years (Table 2.1 EN 1990)
Fire resistance = 1 hr (Sec. 5.7 EN 1992-1-2)
Table 4.1 Exposure classes = XC3
Materials :
Characteristic strength of concrete, f ck = = N/mm2 (Table F.1 EN 206)
3.1.2(3) Characteristic strength of steel, f yk = N/mm2
Unit weight of reinforced concrete = kN/m3 (Table A.1 EN 1991-1)
Assumed: f bar = mm

Design slab panel

Long span, ly = mm l y/l x =

Short span, lx = mm

SLAB THICKNESS
Minimum thickness for fire resistance = mm Table 5.8 EN 1992-1-2
Table 7.4N Estimated thickness considering deflection control, Try,
h = mm h = mm

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Tutorial 9 (Se e Example 7.4 : Two-way Re straine d Slab) page 2/6

Re f. Calculations Output

DURABILITY, FIRE & BOND REQUIREMENTS

Table 4.2 Min. conc. cover regard to bond, C min,b = mm
Table 4.4N Min. conc. cover regard to durability, C min,dur = mm
4.4.1.2 Min. required axis distance for fire resistance Table 5.8 EN 1992-1-2
a = mm
Min. concrete cover regard to fire,
C min = a - f bar/2 = = mm

4.4.1.3 Allowance in design for deviation, D C dev = mm

4.4.1.1(2) Nominal cover, Use:

C nom = C min + DCdev = mm C nom mm

ACTIONS
Slab selfweight = = kN/m2
Permanent load (Excluding selfweight) = kN/m2
Characteristic permanent action, g k = kN/m2
Characteristic variable action, q k = kN/m2
Design action, n d = 1.35g k + 1.5q k = kN/m2 Table A1.2B : EN 1990

ANALYSIS
ly /lx =
Case : Table 3.14 : BS 8110
Be nding mome nts,

Short span:
M sx1 = b sx1 n d l x2 = kNm/m
M sx2 = b sx2 n d l x2 = kNm/m

Long span:
M sy1 = b sy1 n d l x2 = kNm/m
M sy2 = b sy2 n d l x2 = kNm/m

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Tutorial 9 (Se e Example 7.4 : Two-way Re straine d Slab) page 3/6

Re f. Calculations Output

MAIN REINFORCEMENT
Effective depth,
d x = h -C nom -0.5f bar mm
d y = h -C nom -1.5f bar mm

9.2.1.1 Minimum and maximum reinforcement area, Secondary bar :

A s,min = 0.26(f ctm/f yk) bd = bd
= bd = mm2 /m
A s,max = 0.04Ac = mm2 /m

Short span :
- Midspan , M sx =
6.1 K = M / bd 2f ck
=
=

z = d [ 0.5 + 0.25 - K /1.134) ]

A s = M / 0.87 f yk z Use:
=
=

- Support , M sx =
6.1 K = M / bd 2f ck
=
=

z = d [ 0.5 + 0.25 - K /1.134) ]

A s = M / 0.87 f yk z Use:
=
=

Long span :
- Midspan , M sy =
6.1 K = M / bd 2f ck
=
=

z = d [ 0.5 + 0.25 - K /1.134) ]

A s = M / 0.87 f yk z Use:
=
=

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Tutorial 9 (Se e Example 7.4 : Two-way Re straine d Slab) page 4/6

Re f. Calculations Output

- Support , M sy =
6.1 K = M / bd 2f ck
=
=

z = d [ 0.5 + 0.25 - K /1.134) ]

A s = M / 0.87 f yk z Use:
=
=

SHEAR
She ar force , Table 3.15 : BS 8110

Vsx2
Vsy1 Vsy2
Vsx1

Short span:
V sx1 = b vx1 n d l x = kN/m
V sx2 = b vx2 n d l x = kN/m
Long span:
V sy1 = b vy1 n d l x = kN/m
V sy2 = b vy2 n d l x = kN/m

Design shear force, V Ed =

6.2.2 Design shear resistance,
V Rd,c = [ 0.12 k (100r 1 f ck)1/3 ] bd

k = 1 + (200/d )1/2 2.0 < = 2.0

= Use :
r 1 = A sl/bd 0.02
= Use :
V Rd,c =
=
V min = [ 0.035k 3/2f ck1/2 ] bd
=
=

So, V Rd,c = V Ed

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Tutorial 9 (Se e Example 7.4 : Two-way Re straine d Slab) page 5/6

Re f. Calculations Output

7.4 DEFLECTION
Percentage of required tension reinforcement,
r = A s,req / bd
= =
Reference reinforcement ratio,
r o = (f ck) 1/2 x 10-3 =

Table 7.4N Factor for structural system, K =

r < ro Use equation (1)

l r r
3/ 2

K 11 1.5 fck o 3.2 fck o 1 (1)

d r r

=
Modification factor for span,span less than 7 m
=
Modification factor for steel area provided,
= A s,prov/A s,req =
Therefore allowable span-effective depth ratio,
(l /d )allowble = =
Actual span-effective depth
(l /d )actual = (l /d )allowble

CRACKING
7.3.3 h = mm
9.3.1 Main bar :
S max,slabs = 3h 400 mm = mm
Max. bar spacing = mm < S max,slabs
Secondary bar :
S max,slabs = 3.5h 450 mm = mm
Max. bar spacing = mm < S max,slabs

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Tutorial 9 (Se e Example 7.4 : Two-way Re straine d Slab) page 6/6

Re f. Calculations Output

DETAILING

msy '09