Contents Italian NTC 2008 RC PN 001

User Manual: Italian NTC 2008 RC-PN-001

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Software Verification
PROGRAM NAME:
REVISION NO.:

SAFE
0

EXAMPLE Italian NTC 2008 RC-PN-001
Slab Punching Shear Design
PROBLEM DESCRIPTION
The purpose of this example is to verify slab punching shear design in SAFE
The numerical example is a flat slab that has three 8-m spans in each direction, as
shown in Figure 1.
0.3 m

A

B
8m

C
8m

D 0.3 m
8m

0.6 m

4
0.25 m thick flat slab
8m

3
Columns are 0.3 m x 0.9 m
with long side parallel
to the Y-axis, typical

8m

Concrete Properties
Unit weight = 24 kN/m3
f'c = 30 N/mm2

2
8m

Y
X

1

Loading
DL = Self weight + 1.0 kN/m 2
LL = 4.0 kN/m2

0.6 m

Figure 1: Flat Slab for Numerical Example

The slab overhangs beyond the face of the column by 0.15 m along each side of
the structure. The columns are typically 0.3 m x 0.9 m with the long side parallel
to the Y-axis. Thick plate properties are used for the slab.
The concrete has a unit weight of 24 kN/m3 and a f 'c of 30 N/mm2. The dead load
consists of the self weight of the structure plus an additional 1 kN/m2. The live
load is 4 kN/m2.

EXAMPLE Italian NTC 2008 RC-PN-001 - 1

Software Verification
PROGRAM NAME:
REVISION NO.:

SAFE
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TECHNICAL FEATURES OF SAFE TESTED
 Calculation of punching shear capacity, shear stress and D/C ratio.
RESULTS COMPARISON
Table 1 shows the comparison of the punching shear capacity, shear stress ratio
and D/C ratio obtained from SAFE with the punching shear capacity, shear stress
ratio and D/C ratio obtained by the analytical method. They match exactly for
this problem.
Table 1 Comparison of Design Results for Punching
Shear at Grid B-2
Method

Shear Stress
(N/mm2)

Shear Capacity
(N/mm2)

D/C ratio

SAFE

1.100

0.578

1.90

Calculated

1.099

0.578

1.90

COMPUTER FILE: ITALIAN NTC 2008 RC-PN-001.FDB
CONCLUSION
The SAFE results show an acceptable comparison with the independent results.

EXAMPLE Italian NTC 2008 RC-PN-001 - 2

Software Verification
PROGRAM NAME:
REVISION NO.:

SAFE
0

HAND CALCULATION
Hand Calculation for Interior Column using SAFE Method
d=

[( 250 − 26 ) + ( 250 − 38)]

2 = 218 mm

Refer to Figure 2.
u 1 = u = 2•300 + 2•900 + 2•π•436 = 5139.468 mm
1172

Note: All dimensions in millimeters
Critical section for
punching shear
shown dashed.

Y
436

150 150 436
B

A

Column

436

Side 3

Side 1

Side 2

450

X

1772

450

Center of column is
point (x1, y1). Set
this equal to (0,0).

436

Side 4
D

C

Figure 2: Interior Column, Grid B-2 in SAFE Model
From the SAFE output at Grid B-2:
V Ed = 1112.197 kN
k 2 M Ed2 = 41.593 kN-m
k 3 M Ed3 = 20.576 kN-m

EXAMPLE Italian NTC 2008 RC-PN-001 - 3

Software Verification
SAFE
0

PROGRAM NAME:
REVISION NO.:

Maximum design shear stress in computed in along major and minor axis of column:
VEd  k2 M Ed ,2u1 k3 M Ed ,3u1 
+
vEd =
1 +

ud 
VEdW1,2
VEdW1,3 

(EC2 6.4.4(2))

c2
W1 = 1 + c1c2 + 4c2 d + 16d 2 + 2π dc1
2
W1,2
=

9002
+ 300 • 900 + 4 • 300 • 218 + 16 • 2182 + 2π • 218 • 900
2

W1,2 = 2,929, 744.957 mm2

3
W
=
1,3

9002
+ 900 • 300 + 4 • 900 • 218 + 16 • 2182 + 2π • 218 • 300
2

W1,2 = 2, 271,104.319 mm2

VEd  k2 M Ed ,2u1 k3 M Ed ,3u1 
vEd =
+
1 +

ud 
VEdW1,2
VEdW1,3 

1112.197 • 103 
41.593 • 106 • 5139.468
20.576 • 106 • 5139.468 
1
vEd =
+
+


5139.468 • 218  1112.197 • 103 • 2929744.957 1112.197 • 103 • 2271104.319 
vEd = 1.099 N/mm2
Thus v max = 1.099 N/mm2
C Rd ,c = 0.18 γ c = 0.18/1.5 = 0.12

(EC2 6.4.4)

The shear stress carried by the concrete, V Rd,c , is calculated as:
13
VRd ,c C Rd ,c k (100 ρ1 fck ) + k1σ cp 
=

(EC2 6.4.4)

with a minimum of:
v=
Rd ,c

(v

min

k=
1+
k1

+ k1σ cp )

200
≤ 2.0 = 1.9578
d

= 0.15.

EXAMPLE Italian NTC 2008 RC-PN-001 - 4

(EC2 6.4.4)
(EC2 6.4.4(1))
(EC2 6.2.2(1))

Software Verification
PROGRAM NAME:
REVISION NO.:

ρ1 =

SAFE
0

As1
≤ 0.02
bw d

Area of reinforcement at the face of column for design strip are as follows:
A s in Strip Layer A = 9204.985 mm2
A s in Strip Layer B = 8078.337 mm2
Average A s = ( 9204.985 + 8078.337 ) 2 = 8641.661 mm2

ρ 1 = 8641.661 ( 8000 • 218 ) = 0.004955 ≤ 0.02

ν min = 0.035k 3 2 f ck 1 2 = 0.035 (1.9578 )

3/2

vRd ,c = 0.12 • 1.9578 (100 • 0.004955 • 30 )

( 30 )

13

Shear Ratio
=

v max
=
vRd ,c

1/2

= 0.525 N/mm2

+ 0  = 0.5777 N/mm2

1.099
= 1.90
0.5777

EXAMPLE Italian NTC 2008 RC-PN-001 - 5



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