Homework Project - Steel and Timber Design

University of Namibia: Faculty of Engineering and Information Technology

Design of Steel and Timber Structures: Homework Project – Design of Concrete

Elements

Prof L. Haefner

Group Members: Louw Z.V.

Problem Statement: Position 1

Design a 2 – span reinforced continuous slab, according to the relevant standards, which is

meant to withstand the given loading conditions:

Loads:

Self-weight = 25kN/m3

Screed = 1kN/m2

Imposed Loads = 5kN/m2

Relevant standards:

Euro code 2; Loads/Actions

Euro code 2; Design of Concrete

The final design should be done for Ultimate Limit State (ULS) and Serviceability Limit

State (SLS) loading conditions.

Material Data

Concrete: C25/30; Exposure class XC2; fcd = α . f ckγc

= 0,85 .25

1,5 = 14,17MPa

Steel: Class A 500 steel; fyd = f ykγ s

= 5001,15

= 435MPa

Slab dimensions: Plate thickness = 18cm

Width of a single slab = 5m

Span of both slabs = 14m

Design Loads

Concrete Slab self – weight = 0,18m*25kN/m3*1m = 4,5kN/m

Total weight of concrete and slab = 4,5kN/m + 1kN/m2*1m = 5,5kN/m

Partial Safety Factors for ULS: Permanent loads; γg = 1,35

Imposed loads; γq = 1,5

Design loads for ULS: qed = gk*1,35 + qk*1,5 = Gk + Qk

= 5,5kN/m*1,35 + 5kN/m*1,5

= 14,925kN/m

Partial Safety Factors for SLS: Permanent loads; γg = 1,00

Imposed loads; γq = 1,00

Design loads for SLS: qed = 5,5kN/m + 5kN/m

= 10,5kN/m

Loading Cases: To be applied to both SLS and ULS

Load Case 1: Factored permanent loads

Load Case 2: Factored imposed loads

Load Case 3: Factored imposed loads on one span only

These load cases in various combinations of permanent and imposed loads will result in

bending moments within the slab, the analysis will focus on the design of the slab for the

maximum moment induced by these load combinations.

Design moments and required area of steel

The design moments will be taken from the structural analysis software TwoDFrame, this

software will also be used to find the area of reinforcement required for the design. However,

for the sake of clarity, a sample calculation detailing how the bending moment as well as the

required amount of steel is found will be provided. The sample calculation will be prepared

for loading cases 1 and 3 to find the maximum bending moment in this section to be used for

bottom reinforcement as well as the actual required amount of steel. The sample calculation

will then be compared to the TwoDFrame analysis.

Sample Calculation:

MEd = k*Gk(Qk)*L2; k = 0,07 for load case 1 and k = 0,096 for load case 2; L = 5m;

Gk = 7,425kN/m; Qk = 7,5kN/m

MEd = (0,07*7,425kN/m*25m2) + (0,096*7,5kN/m*25m2)

MEd = 30,994kNm

As,bott. = ω * b * d * f cdσsd

+ N Ed

σsd ; ω is obtained from the EC2 tables; b = 1m; d = effective

depth to be determined from an assumed reinforcement diameter; fcd = 14,17MPa; σsd to be

obtained from the relevant EC2 tables; NEd = 0.

To use the EC2 table from which we obtain ω and σsd we must first calculate µEds.

µEds = M Ed

b .d2 . f cd ; MEd = 30,994kNm; b = 1m; fcd = 14,17MPa

d = h – c – 12

ϕ ; h = 180mm; c – from exposure class XC2 = 35mm + 10mm = 45mm;

ϕ = 14mm (assumed).

d = 180mm – 45mm – 7mm = 128mm

µEds = 0,030994MNm

1m∗(0,128m)2∗14,17MPa = 0,13

ω = 0,1401; σsd = 448,6MPa

As,bott. = (0,1401 )∗(1m )∗(0,128m )∗(14,17MPa)

448,6MPa = 5,66cm2

The sample calculation can now be compared to the TwoDFrame analysis:

As can be seen from the TwoDFrame analysis, the maximum bending moment for the section

N01 – N02 is 30,81kNm as compared to 30,994kNm which was found in the sample

calculation.

For the required area of reinforcing steel in section N01 – N02, the following values were

obtained using the TwoDFrame analysis for ULS:

As can be seen from the diagram, the area of steel required for bottom reinforcement As(Bot)

is given as 5,61cm2 as compared to the calculated value of 5,66cm2. This shows that the

TwoDFrame analysis can be used for finding the maximum moment due to the various

loading cases as well as finding the required area of reinforcing steel.

Selection of reinforcing fabrics

It is required for the design that a steel fabric be used; to decide which fabric to use we need

only consider the required area of steel as given by the TwoDFrame analysis. The analysis

was done to find the highest bending moment induced by the given loading; this is used to

calculate the area of steel per meter. The results for the ULS and SLS designs will be

presented separately.

ULS design:

Load cases 1 and 2 – Bending moment:

Load cases 1 and 2 – Area of reinforcement, top and bottom:



The TwoDFrame analysis shows that for the bottom reinforcement we need 5,61cm2/m in

sections N01 – N02 and N02 – N03, and that at N02 we require 8,83cm2/m of reinforcement

to resist the tensile forces induced in the concrete as a result of the external loading.

SLS design





The SLS design indicates a maximum bending moment in sections N01 – N02 and N02 –

N03 of 21,50kNm which requires 3,83cm2/m of steel reinforcing. The maximum bending

moment at N02 is given as -32,81kNm which requires 6,00cm2/m of steel reinforcing.

Selecting the required reinforcing fabric

ULS consideration:

Bottom steel: Area of steel required = 5,61cm2/m

Given this condition we select the following steel fabrics:

R335A – Starting at N01 and spanning the entire 5m up to N02. This is the same for section

N02 – N03.

This does not satisfy the amount of reinforcement required for the design, we therefore add a

second steel fabric but in order to economise we select a length not equal to that of the

section N01 – N02 or N02 – N03. We instead select the following:

R257A – Starting a distance 0,472m from N01 and ending 3,4m away. The length of this

reinforcement is found by considering the length in the span within which the bending

moment in the beam is higher than half the maximum bending moment. This length was

found as 2,8m and for safety we add the effective depth to both ends to come to a total length

of 3,056m for the fabric.

Top Steel: Area of steel required = 8,83cm2/m

Chosen steel fabric: R524A centred above N02 and extending 1,5m to both sides.

To select the second fabric we follow the same procedure as for the bottom steel, we select an

R425A fabric centred above N02 and extending 0,7m to both sides.

SLS consideration:

Bottom steel: Area of steel required = 3,83cm2/m

For this we select the following steel fabrics:

R257A starting at N01 and continuing until N02 and starting at N02 and ending at N03

R188A starting 0,465m from N01 and going on for 2,8m and the same for N02 – N03

Top Steel: Required area of steel = 6cm2/m

The following steel fabrics will be used for this reinforcement:

R424A centred above N02 and extending 1,5m to each side.

R257A centred above N02 and extending 1,13m to both sides of N02.

Reinforcement for hole in slab

ULS considerations:

Longitudinal steel requirements:

Top steel area required: 8,83cm2/m

Amount if steel in hole: 8,83cm2/m * 1,5m = 13,245cm2

Selected rebars: 4 ϕ20 bars; 2 bars parallel to each of the 1,2m sides of the hole with lengths

equal to 1,2m and an added 0,5m extending beyond each of the 1,5m long sides of the hole;

spacing between first bar and edge of hole is 2cm and the same distance is maintained to the

next bar.

Bottom steel area required: 5,61cm2/m

Amount of steel passing through hole = 5,61cm2/m * 1,5m = 8,42cm2

Rebars selected: 4 ϕ16 bars; located the same as the top steel with a spacing of 2cm between

the edge of the hole and the first bar and the same spacing to the next bar.

Transverse steel requirements:

Top steel area: 4,02cm2/m

Steel passing through hole: 4,02cm2/m * 1,2m = 4,8cm2

Rebars selected for reinforcement: 4 ϕ12 bars; 2 on each of the sides parallel to the central

bar extending 0,5m beyond the edge of the hole with a spacing of 2cm between the first bar

and the edge of the hole and also between the first and second bars.


Steel passing through the hole: 2,26cm2/m * 1,2m = 2,71cm2

Selected rebar: 2 ϕ14 bars placed exactly as the top transverse steel.

SLS considerations:

Longitudinal steel requirements:

Top steel area required: 6,00cm2/m

Amount if steel in hole: 6,00cm2/m * 1,5m = 9cm2

Selected rebars: 6 ϕ14 bars; 3 bars parallel to each of the 1,2m sides of the hole with lengths

equal to 1,2m and an added 0,5m extending beyond each of the 1,5m long sides of the hole;

spacing between first bar and edge of hole is 1,5cm and the same distance to, and between,

the next two bars.


Amount of steel passing through hole = 3,83cm2/m * 1,5m = 5,75cm2

Rebars selected: 4 ϕ14 bars; located the same as the top steel with a spacing of 2cm between

the edge of the hole and the first bar and the same spacing to the next bar.

Transverse steel requirements:

Top steel area: 3,14cm2/m

Steel passing through hole: 3,14cm2/m * 1,2m = 3,77cm2

Rebars selected for reinforcement: 4 ϕ12 bars; 2 on each of the sides parallel to the central

bar extending 0,5m beyond the edge of the hole with a spacing of 2cm between the first bar

and the edge of the hole and also between the first and second bars.


Steel passing through the hole: 2,26cm2/m * 1,2m = 2,71cm2

Selected rebar: 2 ϕ14 bars placed exactly as the top transverse steel.

Check for shear reinforcement

We follow the procedure for determining whether or not the slab needs shear reinforcement:

Find VEd1 – shear force at the support = VA - qEd . a

2 = 27,345kN

Find υEd = V Ed1

0,9 . d . b = 231,934kN/m2

Find υRd,c from the provided table using ρl = 0,452%

Through interpolation, υRd,c = 0,547MPa

Multiply υRd,c by the modification factor of 1,00 and compare with υEd1. If υEd < υRd,c

the slab does not require any shear reinforcement.

231,934kN/m2 < 547kN/m2 the slab is therefore safe without any shear reinforcement.

This is a check for shear reinforcement at N01 under ULS consideration.

We now check for shear reinforcement at N02:

VEd1 = 44,77kN

υEd = 379,73kN/m2

υRd,c = 508,64kN/m2

υEd < υRd,c – 379,73 < 508,64

We can therefore see that no shear reinforcement is needed for ULS considerations.

Check for shear reinforcement for SLS at N01

VEd1 = 18,63kN

υEd = 157,37kN/m2

υRd,c = 490kN/m2 using ρl = 0,25%

υEd < υRd,c – 157,37 < 490

Check for SLS shear reinforcement at N02

VEd1 = 30,185kN

υEd = 256kN/m2

υRd,c = 285.84kN/m2 with ρl = 0,378%

υEd < υRd,c – 256 < 285,84

SLS design also does not require any shear reinforcement.

Documents

Homework Project - Steel and Timber Design