answer for question 4 structural design

EBVS 4203 STRUCTURAL DESIGN 2013

ANSWER FOR QUESTION 1

GRADE 35 REIFORCED CONCRETE

(a) BEAM SELF WEIGHT = 24 kN/m3 x 0.5 m x 0.3 m = 3.6 kN/m

(b) Ultimate Design Load,

TOTAL DEAD LOAD = (20 + 3.6) kN/m = 23.6 kN/m

Design Load, w = 1.4 gk + 1.6 qk = (1.4 x 23.6) + (1.6 x 10) = 49.04 kN/m

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500 mm

300 mm

5 m SPAN


(c) Effective Depth, d = 460 mmfy = 460 N/mm2

Steel bar size = 20 mm

M = wl2/8 = 49.04 x (5) 2 8 = 153.25 kNm

Mu = 0.156fcubd2

= 0.156(30)(300)(460)2

= 297.09 kNm

Mu > M : Therefore, design as a singly reinforced beam

TENSION REINFORCEMENT

As = M 0.95 fy z

K = M = 153.25 x 106

fcubd2 (30) (300) (460)2

= 0.080

Z = d [0.5 + (0.25 – K/0.9)]

= 460 [0.5 + (0.25 – 0.080/0.9)]

= 414.68mm

BS 8110 clause 3.4.4.1 limits z not exceed 0.95d

0.95d = 0.95(460) = 437mm

Therefore, z < 0.95d (OK)

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As = M 0.95 fy z

= 153.25 x 106

0.95 x 460 x 414.68 = 845.68 mm2

Using the steel area table,

we choose 3T20 (942 mm2) that mean number of bars required for the main reinforcement is 3 bars.

(d) THE REQUIRED SPACING OF THE LINKS (8 mm; Asv = 100mm2)

Sv < Asv (0.95fyv) 0.4b

Sv < 100 (0.95 x 250) 0.4 x 300

Sv < 197.92mm

BS8110 limits the maximum spacing of links to not greater than 0.75d. 0.75 x 460 = 345mm

We assume that required spacing is 300mm c/c R8 @ 300mm c/c

(e) DEFLECTION CRITERIA CHECK

3


M/bd2 = 153.25x106 / (300)(460)2

= 2.414

fs = 2 fyAs,reqd x 1

3 As,provd βb

fs = 2/3 x 460 x (846/942) = 275 N/mm2

From Table 3.7, MF = 1.06

ALLOWABLE L/d is 1.06 x 20 = 21.2

ACTUAL L/d is 5000/460 = 10.9

THE ACTUAL L/d < THE ALLOWABLE L/d

THEREFORE DEFLECTION OK


4

3T20

R08 at 300 c/c

500 mm

300 mm


COLUMN DESIGN

DEAD LOAD = 1800 kN

IMPOSED LOAD = 1000 kN

CLEAR HEIGHT = 4.0m

(a) DETERMINATION OF SHORT OR SLENDER COLUMN.

From Table 5.1, β = 1.00Lex = β lox = 1.00 x 4000 = 4000mmLey = β loy = 1.00 x 4000 = 4000mm

SLENDERNESS RATIOS

Lex/h = 4000/500 = 8.0 (< 15)

Ley/h = 4000/300 = 13.3 (< 15)

THEREFORE, DESIGN AS SHORT COLUMN.

(b) THE LIKELY FAILURE MODE FOR THIS COLUMN.

The failure mode is crushing of the concrete and subsequently the steel reinforcement due to compression.

5

500 mm

300 mm


(c) STEEL BAR = 25mm

DESIGN LOAD = P = 1.4 Gk + 1.6 Qk

= 1.4(1800) + 1.6(1000)

= 4120 kN

MAIN REINFORCEMENT

N = 0.4 fcuAc + 0.8 fyAsc

4120 x 103 = 0.4 x 40 x ((300 x 500) - Asc) + 0.8 x 460 x Asc

Asc = 4886.36 mm2

From the steel table, the suitable size and number for main reinforcement is 10 bars of 25mm diameter with total area of 4908mm2.

THEREFORE, PROVIDE 10T25 (Asc = 4908mm2)

(d)

LINKS

Diameter of links is one-quarter times the diameter of the largest main reinforcement, but not less than 8mm,

¼ x 25mm = 6.25mm

Provide 8mm diameters mild steel for links.

Spacing for links is the lesser of:

12 x 25 = 300mm

Smallest cross-sectional dimension of the column = 250mm

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(e)


7

10T25

R08 at 250 c/c

500 mm

300 mm


(a) The Ultimate Limit State is the state where a structure or part of a structure reaches its

ultimate strength, where ruptures to critical sections occur. This will lead to instability of

the structure and eventually collapse of critical parts.

The Ultimate Limit State condition is computationally checked at a certain point along

the behavior function of the structural scheme, located at the upper part of its elastic zone

at approximately 15% lower than the elastic limit. That means that the Ultimate Limit

State is a purely elastic condition, located on the behavior function far below the real

Ultimate point which is located deeply within the plastic zone. The rational for choosing

the Ultimate Limit State at the upper part of the elastic zone is that as long as the

Ultimate Limit State design criteria is fulfilled, the structure will behave in the same

way under repetitive loadings, and as long as it keeps this way, it proves that the level of

safety and reliability assumed as the basis for this design is properly maintained and

justified, (following the probabilistic safety approach).

(b) Partial safety factors are basic indicators, which determine structural dimensions in

relation to loading. The probabilistic assessment of reliability is performed as a

parametric study in the first part of the numerical analysis. The probability of failure is

analyzed in dependence on values of partial safety factors of material, permanent loading

and longtime variable loading.

It is introduced to take account of:

The possibility of increases in load.

Inaccurate assessment of load effects.

Unforeseen stress redistribution.

Variation in dimensional accuracy.

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(c)

(i) TOTAL UNFACTORED DEAD LOAD (in kN/m) ON THE BEAM.

BRICKWORK = WEIGHT DENSITY x THICKNESS x HEIGHT = 22 x 0.230 x 2

= 10.12 kN/m

CONCRETE = WEIGHT DENSITY x BEAM’S SIZE = 24 x 0.25 x 0.50 = 3.00 kN/m

TOTAL UNFACTORED DEAD LOAD = 10.12 kN/m + 3.00 kN/m = 13.12 kN/m

(ii) BEAM REACTION, MAXIMUM SHEAR FORCE AND MAXIMUM BENDING MOMENT

9

500 mm

4.0 m

250 mm

2.0 m

230 mm

13.12 kN/m


ΣMA = 0

ΣMA = (13.12 kN/m x 4 m x 2 m) - (RB x 4 m)

RB = 26.24 kN

ΣPV = 0

RA – (13.12 kN/m x 4 m) + RB = 0

RA = 26.24 kN

Mmax = 26.24 kN x 2 m

= 52.48 kNm

(iii)

SHEAR FORCE DISTRIBUTIONS

10

26.24 kN


Shear Force Diagram

BENDING MOMENT DISTRIBUTIONS

Bending Moment Diagram


SLAB DESIGN

11

26.24 kN

2 m

52.48 kNm

130 mm


DEAD LOAD (INCLUDING SELF-WEIGHT OF SLAB) = 5.0 kN/m2

IMPOSED LOAD = 2.0 kN/m2

fcu = 30 N/mm2

fy = 460 N/mm2

(a) Ultimate design load.

Design load, n = (1.4 x gk) + (1.6 x qk)= (1.4 x 5.0) + (1.6 x 2.0)= 10.2 kN/m2

(b) Two-way slabs verification.

Ly/Lx = 4.2/3.5 = 1.2 < 2 ( Slab is design as two-way slab)

(c) Positive moments at mid-span in both directions.

β sx = 0.074β sy = 0.056

12


mid-span moment in the short span: m sx = β sx n l x

2

= 0.074 x 10.2 x (3.5)2

= 9.25 kNm/m

mid-span moment in the long span: m sy = β sy n l x

2

= 0.056 x 10.2 x (3.5)2

= 7.00 kNm/m

(d) Main bar, Ф= 10mmEffective depth for main bars = h – (Ф/2 + cover) = 130 – (5 + 20)

= 105mm

Secondary bar, Ф= 10mmEffective depth for secondary bars = h – (Ф + Ф/2 + cover) = 130 – (10 + 5 + 20)

= 95mm

MAIN REINFORCEMENTK = msx/fcubd2 = 9.25x106/ (30) (1000) (105)2

= 0.028

Z = d [0.5 + (0.25 – K/0.9)]

= 105 [0.5 + (0.25 – 0.028/0.9)]

= 101.62mm < 0.95d (OK)

As = Msx

0.95 fy z

= 9.25 x 106

0.95 x 460 x 102

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= 207.62 mm2

PROVIDE T10 AT 250 CENTRES (As = 314mm2/m) IN THE DIRECTION OF SHORT SPAN.

SECONDARY REINFORCEMENTK = msx/fcubd2 = 7.00x106/ (30) (1000) (95)2

= 0.026

Z = d [0.5 + (0.25 – K/0.9)]

= 95 [0.5 + (0.25 – 0.026/0.9)]

= 92.17mm < 0.95d (OK)

As = Msx

0.95 fy z

= 7.00 x 106

0.95 x 460 x 92 = 374.11 mm2

PROVIDE T10 AT 200 CENTRES (As = 392mm2/m) IN THE DIRECTION OF LONG SPAN.

(e) SPAN/EFFECTIVE DEPTH RATIO CHECK

M/bd2 = 9.25x106/ (1000)(105)2

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= 0.839

fs = 2 fyAs,reqd x 1

3 As,provd βb

fs = 2/3 x 460 x (208/314) = 203 N/mm2

From Table 3.7, MF = 1.86

ALLOWABLE L/d is 1.86 x 26(CONTINUOUS) = 48.36

ACTUAL L/d is 4200/105 = 40

THE ACTUAL L/d < THE ALLOWABLE L/d

THEREFORE DEFLECTION OK

REFERENCES

EBVS4203/EDVS3203 STRUCTURAL DESIGN OUM

REINFORCED CONCRETE DESIGN THEORY AND EXAMPLE, second edition, T.J.MacGinly and B.S.Choo

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BSI British Standards Publications

www2.iccsafe.org/.../07_PDFs/Chapter%2016_Structural%20Design.pdf

http://webstore.ansi.org/FindStandards.aspx?Action=displaydept&DeptID=3176&Acro=BSI&DpName=BSI:%20British%20Standards%20Institution&source=msn&adgroup=bsi1

http://www.standardsuk.com/

http://www.astm.org/

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answer for question 4 structural design

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