Topic 2 Loadings.pdf

Structural Idealization and Loading Structural Idealisation This generally means breaking the complete structure down into single elements that can be easily analyse and design. It is rarely possible to consider the threedimensional structure in its entirety. Consider a concrete beam carrying a water tank as shown below. For analysis purpose we can neglect the thickness of the members and assume the beam is simply supported. The idealized structure is as shown below represented by line drawing for the beam and arrow for the forces that acted on the beam. By applying the principle of structural analysis, if F is known than we can calculate the reaction(R1 and R2) at the support and the maximum shear force and bending moment in the beam. Eventually we can design the suitable size of the beam that can safely carry the water tank.

F/2

R1 (a) Actual structure

F/2

R2

(b) Idealised structure Idealised Structure

Example A reinforced concrete flat roof with a parapet wall at the end is as shown below. Draw the idealised beam and name the forces that acts on the beam.

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Sectional and Plan View of a structure

Solution

Uniformly distributed load from live load + dead load

Concentrated Load

(slab + beam)

from parapet Support reaction 2.0 m

6.0 m

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Loads on Individual Structural Elements A structural element is a portion of a structure that can be usefully considered as a separate entity. At an appropriate time in the design process a slab, a beam, a wall or a column can be visualised as isolated from the rest of the structure and calculations can be performed on that element to determine a suitable choice of size or section.

EXAMPLE 1 Timber beams spanning 4.0 m and spaced at 3.0 m centres, as shown in Figure below , supported timber floor comprising joists and boards with a plaster ceiling. Other design data:  Self-weight of boards and fl oor joists 0.23 kN/m2  Self-weight of ceiling 0.22 kN/m2  Imposed load on floor 1.50 kN/m2  Self-weight of timber beam 0.6 kN (assumed) A weight for the beam is assumed because at this stage in the design process the size of the beam is not known.

Floor plan

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SOLUTION To determine the uniformly distributed load (UDL), visualise a single beam removed from the structure as shown below: The load on one square metre of floor is multiplied by the area supported, 4.0 m x 3.0 m,(tributary area) to give the total UD load on the beam as follows:

Isolated timber floor beam

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Loads on timber beam Joists and boards Ceiling

0.23 x 4.0 x 3.0 0.22 x 4.0 x 3.0

Timber beam Imposed load Totals

1.5 x 4.0 x 3.0

Dead loads (self-weight) kN

Imposed loads kN

2.76 2.64 0.60 6.0

18.0 18.0

Total UD load for ultimate limit states = 6.0 x 1.35 + 18.0 x 1.50 = 35.1 kN

EXAMPLE 2 Steel floor beams arranged as shown below support a reinforced concrete slab which carries a screed. Calculate the total Uniformly Distributed load on each beam using this design data:  slab thickness =150 mm ● screed weight =1.2 kN/m2 ● imposed load on slabs =5.0 kN/m2 ● mass of steel beams =60 kg/m. \

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SOLUTION Unit weight of reinforced concrete = 25 kN/m2 Concrete slab self-weight = 0.15 x 25 = 3.75 kN/m2 Steel beam self-weight = 60 x 0.0098 = 0.59 kN/m

Loads on steel beam Screed Concrete slab Steel beam Imposed load Totals

1.2 x 6.0 x 4.0 3.75 x 6.0 x 4.0 0.59 x 6.0 5.0 x 6.0 x4.0

Dead loads (self-weight) kN

Imposed loads kN

28.8 90.0 3.5 122.3

120.0 120.0

Total UD load for ultimate limit states = 122.3 x 1.35 + 120.0 x 1.50 = 345.1 kN EXAMPLE 3 Find the beam loads and the reactions transmitted to the walls for the steelwork arrangement shown below using this design data: ● reinforced concrete (RC) slab thickness 100 mm ● screed weight 1.0 kN/m2 ● imposed load on slabs 3.0 kN/m2 ● mass of steel beams 80 kg/m.

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Floor plan SOLUTION Concrete slab self weight = 0.10 x 25 = 2.50 kN/m2 Steel beam self weight = 80 x 9.81/1000 = 0.78 kN/m

UD loads on beam A Screed Concrete slab Steel beam Imposed load Totals

1.0 x 5.0 x 2.5 2.5 x 5.0 x 2.5 0.78 x 5.0 3.0 x 5.0 x 2.5

Dead loads (self-weight) kN

Imposed loads kN

12.5 31.3 3.9 47.7

37.5 37.5

Total UD load for ultimate limit states = 47.7 x 1.35 + 37.5 x 1.50 = 120.6 kN

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As beam A is loaded symmetrically each end reaction is half the total load Loads and reactions are shown in diagram below:

Beam A isolated

Beam B

UD loads on beam B

Steel beam Total

0.78 x 5.0

Dead loads (self-weight) kN

Imposed loads kN

3.9 3.9

-

Total UD load for ultimate limit states = 3.9 x 1.35 =_ 5.3 kN The diagram below shows the loads on beam B and the reactions transmitted to the walls:

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Beam B isolated

EXAMPLE 4 A reinforced concrete mezzanine floor slab is simply supported on steel beams and columns as shown in the Figure below. Calculate the beam and column loads using this design data: ● slab thickness 200 mm ● weight of screed 1.5 kN/m2  mass of steel beams 120 kg/m ● mass of steel columns 100 kg/m ● imposed load on slab 4.5 kN/m2

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Arrangement of mezzanine floor

SOLUTION Concrete slab self-weight = 0.20 x 25 = 5.0 kN/m2 Total self-weight of slab + screed = 5.0 + 1.5 = 6.5 kN/m2 Steel beam self-weight = 120 x 9.81/1000 = 1.18 kN/m Steel column self-weight = 100 x 9.81/1000 = 0.98 kN/m Beam A UD loads on beam A Concrete slab + screed Steel beam

6.5 x 8.0 x 2.5 1.18 x 8.0

Dead loads (self-weight) kN

Imposed loads kN

130.0 9.4

-

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Imposed load

4.5 x 8.0 x 2.5

Totals

-

90.0

139.4

90.0

Total UD load for ultimate limit states = 139.4 x 1.35 + 90.0 x 1.50 = 323.2 kN Reactions for beam A

Beam A isolated

Beam B UD loads on beam B Concrete slab + screed Steel beam Imposed load

Dead loads (self-weight) kN

Imposed loads kN

260.0 9.4 -

180.0

269.4

180.0

6.5 x 8.0 x 5.0 1.18 x 8.0 4.5 x 8.0 x 5.0

Totals

Total UD load for ultimate limit states = 269.4 x 1.35 + 180.0 x 1.50 = 633.7 kN Loads and reactions for beam B

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Beam B isolated Beam C

UD loads on beam C

Steel beam Total

Dead loads (self-weight) kN

Imposed loads kN

11.8 11.8

nil

1.18 x 10.0

Total UD load for ultimate limit states = 11.8 x 1.35 = 15.9 kN Loads and reactions for beam C

Beam C isolated

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Columns Self-weight of one column ULS = 1.35 x 2.4 = 3.2 kN Column One

Total column load for ULS= 166.4 + 161.6 + 3.2 = 331.2 kN

Column Two

Total column load for ULS= 2 x 166.4 + 316.9 + 3.2 = 652.9 kN

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ASSIGNMENT 1 A series of reinforced concrete beams at 5.0-m centres span 7.5 m onto reinforced concrete columns 3.5 m high, as shown in Figure below . The beams, which are 575 mm deep by 250 mm wide, carry a 175-mm-thick reinforced concrete slab, which can be considered as simply supported. The columns are 250 mm by 250 mm in cross-section. The slab carries a screed weighing 1.4 kN/m2 and an imposed load of 3.0 kN/m2 . Find the total UD load on one beam, the reaction to one column and the loads from one column to its foundation.

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TUTORIAL 2 The floor plan of a composite structure (concrete slab plus steel beam) is below. The reinforced concrete slab has an overall thickness of 200 mm and is to be designed to carry an imposed load of 3 kN/m2 including floor finishes and ceiling loads of 1 kN/m2. Calculate the design load(kN/m) acting on steel beam A2 – B2. Assume weight of steel beam is 70 kg/m.

3 3m 6m

2 3m 1

6m

3m

A

B

C

Floor Plan

Solution Assume the shaded region as the tributary area to be supported by beam A2 – B2. Loading Dead Load (kN/m) Slab: 25 x 0.2 x 3 Floor finishes and ceiling: 1 x 3 Beam: 70 x 9.81 x 10-3 Total Imposed Load : 3 x 3

= = = = =

15 kN/m 3 kN/m 0.7 kN/m 18.7 kN/m 9 kN/m

Design load = 1.35gk + 1.5 qk = 1.35 x 18.7 + 9 x 1.5 = 38.75 kN/m

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Idealised beam A2 – B2:

w = 38.75 kN/m

Exercise Q1. What is the difference between dead and live load in a structure and give two example for each type of load.. Q2. Calculate the self-weight of reinforced concrete beam of breadth 300 mm, depth 600 mm and length 6000 mm. Q3. The floor of a classroom is made of 125 mm thick concrete. If the floor is a slab having a length of 8 m and width of 6 m, determine the total design load in kN for dead and live load. Use values in Table of dead and imposed load given above. Q4. Referring to the problem given above, calculate the design load of (i) Beam A1-B1 (ii) Beam A1-A3 (iii) Column A1 (Assume all beams are simply supported) Q5. Name the modes of failure that can occur on a reinforced concrete (i) beam (ii) column.

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DISTRIBUTION OF LOAD ON BEAMS FROM ONE- WAY AND TWO -WAY SLAB

Floor slabs can be designed as either one-way spanning or two-way spanning as shown in Figures (a) and (b)

In the case of one-way spanning slabs the entire load is distributed to the two main beams. Two-way spanning slabs distribute load to main beams along all edges.

The load of the slabs is distributed on the supporting beams as shown below (based on cracked pattern of slab at yield): Ly

Ly

Lx Lx

Two ways Slab

One way Slab

One way slab and two-way slabs can be recognised by dividing Ly(longer size) to Lx(shorter size). When Ly/Lx ≤ 2 then it is considered as two- way slab When Ly/Lx >2 then it is considered as one- way slab

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Given the floor plan as shown below. The types of floor can be determined using the expression above.

The distribution of loads from slabs to supporting beams.

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Figure: The types of loading from slabs distributed to the supported beams

EXERCISE Given the following floor plan, plot the distribution of loading of the floor on the surrounding beams. 5500 mm

2750 mm

A

2750 mm B

3000 mm

3000 mm

D C

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The trapezoidal load, triangular load and rectangular load can be converted to uniformly distributed load for ease of calculations using the following expressions: Trapezoidal load to Uniformly distributed load(kN/m):

Triangular load to Uniformly distributed load(kN/m):

Rectangular load to Uniformly distributed load(kN/m):

Where n = kN/m2

EXAMPLE The figure below shows the plan of a precast concrete building. Determine the design load, bending moment and shear force for beam 1a/D-E and beam E/1-2 (assume as simply supported) if the characteristics variable action on the floor is 3.0 kN/m2 and the characteristics self weight of slab is 4.6 kN/m2. The beam E/1-2 also carries a brick wall of 3 m height. Also given: Weight of bricks: 2.6 kN/m2 Slab thickness : 150 mm

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Solution Beam 1a/D-E

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Self weight of beam = 25 x 0.25 x (0.55 – 0.15) = 2.5 kN/m Design load,

w = 1.35 (9.2 + 2.5) + 1.5 (6) = 22.05 kN/m

Maximum Shear force (reaction) = wL/2 = 22.05 x 3/2 = 33.08 kN Maximum bending moment = wL2/8 = 22.05 x 32/8 = 24.81 kNm

BEAM E/1-2

= 4.6 x 3/6 [ 3 – (3/3.75)2] = 5.43 kN/m

Wqk,1 = 3x3/6 [ 3 – (3/3.75)2] = 3.54 kN/m

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Wqk,2 = nLx/2 = 3 x 3/2 = 4.5 kN/m

Self weight of beam = 25 x 0.25 x (0.55 – 0.15) = 2.5 kN/m Self weight of brick wall = 2.6 x 3 = 7.8 kN/m Design load, w = 1.35(5.43 + 6.9 + 2.5 + 7.8) + 1.5 (3.54 + 4.5) = 42.61 kN/m Point load (reaction from beam 1a/D-E) = 33.08 kN

42.61 kN/m

33.08 kN

3.75m

3.75 m

EXERCISE 1. 2. 3. 4.

Calculate the reaction (Maximum shear Force) Calculate the maximum design moment Sketch the shear force diagram Sketch the bending moment diagram

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ASSIGNMENT 2 Q1 Rajah di bawah menunjukkan pelan lantai tingkat kedua sebuah bangunan pasang siap.. Data berikut diberi, Ketebalan papak konkrit = 125 mm Beban kenaan di atas papak = 3.0 kN/m2 Berat konkrit = 25 kN/m3 Berat dinding bata di atas rasuk = 2.0 kN/m Kemasan lantai = 1.5 kN/m2

A.

Dengan mengandaikan rasuk sekunder D/1-3 sebagai tupang mudah, (i) (ii)

B

Kirakan beban rekabentuk pada rasuk. Kirakan momen rekabentuk.

Kirakan beban rekabentuk danm momen r/b rasuk B1-B4 dalam kN/m

125 450

300 All dimensions in mm Beam D/1-3

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Q2. The floor system of a building consists of a 15 cm thick reinforced concrete slab resting on four steel beams, which in turn are supported by two steel girders as shown below. The cross-sectional areas of the floor beams and the girders are 94.8 cm2 and 337.4 cm2, respectively. Determine the permanent action on the beams CG and DH and the girder AD. Given the unit weight of the construction materials as follows: Reinforced concrete: 25 kN/m3 Structural steel: 77.0 kN/m3

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