2.1

Chapter 2 Reinforced concrete and Loading OBJECTIVE/SCOPE: This chapter explains the use of reinforced concrete as a structural material and the use of partial load factors at limit states design. SUMMARY: The fundamental objectives of reinforced concrete design are discussed at limit states. The stress strain relationship of the materials are covered as well as the behaviour of reinforced concrete. The partial load factors on loading and material is explained and a number of examples are given to illustrate loading at ultimate limit state design. LEARNING OBJECTIVES: To understand the limitations and the use of reinforced concrete. To understand the design principles at ultimate limit state and serviceability limit state design. To be able to calculate loads, bending moments and shear forces correctly using the loading code

and apply partial load factors. 2.1 Introduction It is well known that the Romans produced some type of cement. Pipes manufactured from this cement has been in use for more than a 100 years. The process to manufacture cement was developed in 1824 by Joseph Aspdin. A further development was in 1861, when Joseph Monier patented the first reinforced concrete flower pot after a number of experiments. It was only after 1886 when a German Engineer, Koehnen mastered the first theory of reinforced concrete and produced the first reinforced concrete calculations. Thereafter the development was rapid and in 1928 that Freyssinet patented prestressed concrete. Since the Second Word War reinforced concrete theory on yield load design was developed and in an early seventies the Limit States design method emanated from that. 2.2 Concrete Concrete is a material that can withstand a relatively high compressive strength with ease. Unfortunately concrete is less suited to withstand shear stresses and almost has no tensile resistance.

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In ideal situations concrete can be made to withstand a 70 to 84 MPa compressive stress at 28 days but seldom can resist more than 7 MPa in tension. Concrete with a compressive strength of 25 MPa at 28 days is a good average strength for concrete for normal applications. As a result of this favourable characteristic concrete is specially suited for elements or members subjected to compression., such as columns, arches, etc. The principle that the design of reinforced concrete elements is based on is, mainly to supplement the tensile forces in concrete construction with reinforcing bars to resist the tensile stresses while the concrete resist the compressive forces. The reinforcing used is predominantly steel but in certain cases other material, such as timber (bamboo cane), glass, glass fibre, bras, etc. has been used.. In the theory to follow all deductions are limited to the use of steel as a reinforcing material.. Steel is prone to corrosion when exposed to the atmosphere but when placed in concrete (as a reinforcing medium) the corrosion is negated by the favourable PH in the concrete, that normally is approximately 12 (i.e. an alkaline environment). Concrete and steel together interacts well compositely that in many cases reinforced concrete is considered as a distinctive construction material considered. The expansion coefficients of steel and concrete are similar and therefore the behaviour of the composite material is not affected by mild temperature variations. The maintenance of reinforced concrete structures is relatively low in comparison to other material where corrosion protection, in the form of paint or some or other surface applications is imperative. It is also possible to build a wide range of structures economically using reinforced concrete; local labour is readily available and can be trained easily for concrete applications, material is mostly readily available, plant and equipment used for concrete applications are affordable and if careful consideration is given to detail and constructability. 2.3 Elastic design theory Concrete does not behave completely elastic when subjected to a compressive stress. Within certain limits, normally the permissible stress region, it is considered to be elastic and complies with Hooks Law, i.e. compressive stresses are directly proportionate to deformation in the direction of the stress. Due to the ineffectiveness of concrete to resist axial tensile stresses it is completely ignored in the design theory of reinforced concrete. However, when designing concrete in shear, it is assumed that concrete indirectly resists tensile stresses. In practice resistance is offered to tensile stresses in the tensile zone of a section, but due to the presence of possible micro-cracks being manifesting due to temperature changes, and shrinkage this tensile stresses are not relied upon and is considered as an extra measure of safety. Consider a rectangular section of a beam made up of an homogenic elastic material subjected to a positive bending moment (See Figure 2.1). In the fibres above the neutral axis the fibres are in compression whilst beneath the neutral axis the fibres are in tension. In a homogeneous beam section the neutral axis, i.e. the plane where there is no stresses in the fibres, co-insides with the centroid of the section but in a reinforced concrete section it does not occupy any specific position but it varies with the position and the percentage of steel reinforcing.

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The position of the neutral axis can even vary with the change in loading since with small loading the concrete itself can resist the tensile stresses. If the concrete may not resist any axial tensile stress the tensile stress below the neutral axis has to be displaced to the reinforcing below the neutral axis than can resist tensile stresses. When a reinforced concrete section is subjected to an external load the concrete fibres under tensile stress will begin to crack when the concrete reaches its elastic limit. The magnitude of the cracks are sometimes small and can not be observed by the naked and as they develop more and more tensile forces are transferred to the reinforcing steel until the steel resists all tensile forces. It is also evident that the neutral axis gradually moves downward until it reaches a fixed position prior to failure. It is therefore evident that if the reinforcing steel can be placed further away from the neutral axis the more beneficial it will be. A certain portion of stress is resisted by the concrete between the neutral axis the fibres that have now developed cracks (Figure 2.2), but for normal design it is assumed that all tensile stresses are resisted by the reinforcing steel alone.

The formulae used in the theory for reinforced concrete are based on the assumption that the concrete and steel will not deform beyond their elastic zones.

+ Bending moment diagram

Neutral axis

Figure 1

Section Stress distribution

Compression zone Tension zone

Compression Tension

Hairline cracks

Figure 2 Figure 2.2: Hairline crack distribution in tension zone

Figure 2.1: Elastic stress-distribution of a reinforced concrete beam

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The safety factor (SF) used with reinforced concrete ensures that there is an acceptable margin between the safe design load and the load that will result in failure. This safety is ensured by prescribing the limit of maximum permissible stresses in concrete and steel.

Please note: (a) With failure it is implied that material, i.e. concrete and steel is deformed beyond the elastic zone

and plastic deformation takes place. (b) A safeguard is ensured by reducing the stresses in the design process and to design using a safe

stress that includes a safety factor. 2.4 South African Bureau of Standards Codes of Practice Up until approximately 1980 the British Design Code, CP 114-1948 (a design code based on the permissible stress method) was utilised, thereafter the new SABS 0100-1980 was introduced. SABS 0100-1980 was based on CP110-1972 that was less conservative than CP 114. In the elastic design method in CP114 and the corresponding South African Standard Building Regulations the permissible stresses were limited within the elastic zones, whilst in the limit states design in CP 110 and on which SABS 0100 was based, the material is allowed to reach their theoretical yield stresses. The analysis allows for partial load factors to be used in conjunction with the applied loading to move the emphasis to load factors and to allow only small partial material safety factors to allow for the normal deviation in quality control and supervision during manufacturing. The SABS Code of Practice 0100-1 (1985) is based on BS 8110-1985 (1) that was based on previous SABS and British Codes. The new Codes (SANS 10100-1,2000) (2) has the benefit that the reinforcing steel in the structure will fail first when subjected to overloading, (the deformation of the reinforcing steel in the plastic zone increases whilst the stresses in the concrete zone remain constant), and this gives a timeous and clear warning for evacuation of a building. The elastic method had the disadvantage that a structure was over designed, (in other words it had to much reinforcing steel) and that the concrete failed first when over stressed prior to the steel reaching yield stress, that results in a brittle and instantaneous failure without any prior warning. 2.5 Guideline to SANS 10100 The Design Codes shall be read in conjunction with these notes to master this subject matter. The relevant clauses are indicated in brackets: 2.5.1 Limit states design (Clause 3) The purpose of limit states design is to ensure a certain degree of probability that a structure will remain serviceable in its lifetime. The important limit states criteria are as follows:

loadDesign load Failure

=SF

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2.5.1.1 Ultimate limit state (Clause 3.2.2.) This is when a structure or a section of a structure reaches the point when failure will take place subjected to a given load. 2.5.1.2 Serviceability limit state (Clause 3.2.3.) This is subject to the normal use of the structure and the serviceability depends on a number of factors, i..e.: Deflection (Clause 3.2.3.2) excessive deflection of a structure or an element of a structure can

damage other elements or can negatively influence the appearance of a building. SANS 10100, Table 1 gives permissible standards for deflection to be used in design.

Crack formation (Clause 3.2.3.3) the appearance and durability of a structure can be effected detrimentally by cracks.

Vibration (Clause 3.2.3.4) vibration excessive deflection of a structure or an element of a structure can damage.

Other design considerations (Clause 3.2.4): Fatigue (Clause 3.2.4.1) due to cyclical loading the structure can be damaged. Durability (Clause 3.2.4.2) the life expectancy of a structure must be considered due to

environmental conditions. Fire resistance (Clause 3.2.4.3) the tenants must be in a position to vacate the building prior to

structural damage as a result of fire. Lightning (Clause 3.2.4.4) The reinforcing may be used as a lightning conductor. 2.6 Applications of reinforced concrete 2.6.2 Introduction Reinforced concrete can be used for numerous applications, i.e. for low and high rise residential, commercial and industrial, concrete framing (beams and columns), walls, retaining walls, basement walls, slabs, cooling towers, communication towers, masts, foundations, arches, domes, dams and spillways to name a few. Precast concrete as well prestressed, post-tensioned or pre-tensioned concrete elements and structures will not be covered. Reinforced concrete is a structural material akin to structural, timber and masonry used in appropriate applications will be a viable solution to the structure. 2.6.2 Worlds tallest building and free-standing structure The CN tower in Toronto, Canada is the worlds tallest building and free-standing structure (Figure 2.3). This structure is a reinforced concrete structure; remember 1 ft = 0,3048 m and there are 12 inches () in a ft:

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At a height of 553.33 m, it is the world's tallest building and free-standing structure, an important telecommunications hub, and the centre of tourism in Toronto. Each year, approximately 2 million people visit the world's tallest building to take in the breathtaking view and enjoy all of the attractions the CN Tower has to offer.

After 40 months of construction, the CN Tower was opened to the public on June 26, 1976. Today, it is the centre of telecommunications for Toronto serving 16 Canadian television and FM radio stations, the workplace of 550 people throughout the year, and one of Toronto's premier entertainment destinations.

The 1960s ushered in an unprecedented construction boom in Toronto transforming a skyline characterized by relatively low buildings into one dotted with skyscrapers. These buildings caused serious communications problems for existing transmission towers which were simply not high enough to broadcast over the new buildings. Signals bounced off the buildings creating poor television and radio reception for residents. With its microwave receptors at 338 m and 553.33m antenna, the CN Tower swiftly solved the communications problems with room to spare and as a result, people living in the Toronto area now enjoy some of the clearest reception in North America.

Building the CN Tower involved 1,537 workers who worked 24 hours a day, five days a week for 40 months to completion. Tower construction crews moved in on February 6, 1973, and started to remove over 56 metric tonnes of earth and shale for the foundation. Once the foundation was ready, work began on the Tower's 335 m concrete shaft, a hexagonal core with three curved support arms. This involved pouring concrete into a massive mold or "slipform".

Eight months later, the Tower's concrete shaft was the tallest structure in Toronto and by February 1974, it was the tallest in Canada. In August 1974, work began on the seven story tower sphere that would eventually house the observation decks and revolving restaurant. The Tower approached completion in March 1975, when Olga, the giant Russian Sikorsky helicopter flew into the city to lift the 44 pieces of the antenna into place. The Tower was finished on April 2, 1975, and opened to the public June 26, 1976.

When the 44th and final piece of the Tower's antenna was bolted into place April 2, 1975, the CN Tower joined the ranks of 17 other great structures that had previously held the title of World's Tallest Free-Standing Structure. Ross McWhirter, editor of the Guinness Book of World Records, was on hand to record the milestone for history and since then, the Tower has received numerous mentions in the famous book including the World's Longest Metal Staircase and most recently, the World's Highest Wine Cellar. In 1996, the CN Tower's classification was officially changed to the World's Tallest Building and Free-Standing Structure.

In 1995, the CN Tower was classified as one of the seven wonders of the modern world by the American Society of Civil Engineers. The World's Tallest Building shares this designation with the Itaipu Dam on the Brazil/Paraguay border, the Golden Gate Bridge in San Francisco, the Panama Canal, the Chunnel under the English Channel, the North Sea Protection Works off the European coast, and the Empire State Building.

CN Tower - frequently asked questions

Is the CN Tower the World's Tallest Building? Yes, it is recognized by the Guinness Book of World Records as the world's tallest building and

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free-standing structure. The Tower holds a number of other world records including world's longest metal staircase and world's highest wine cellar.

How much did it cost to build the World's Tallest Building? The CN Tower was built for $63 million.

How long did it take to build the CN Tower? It took approximately 40 months to complete the CN Tower. Construction started on February 6,

1973, and wrapped up in June of 1976.

What is the mass of the Tower? The world's tallest building weighs 'in at 117,910 metric tonnes. This is about the equivalent of

23,214 large elephants.

How much concrete is in the Tower? There is a full 40,524 cubic meters of concrete in the CN Tower. There is also 129 km of post tensioning steel. 4 535 metric tones of reinforcing steel, and 544 metric tones of structural steel.

The tower has a glass floor, is it safe? Yes. The Tower's Glass Floor, opened on June 26, 1994, is 23.8 square meters of solid glass and

is five times stronger than the required weight bearing standard for commercial floors. It could withstand the weight of 14 large hippos.

How many elevators arc there in the CN Tower? There are six glass-fronted, high-speed elevators. They travel at a rate of 22 km/hour hour. It takes

only 58 seconds to reach 346m.

Does the Tower sway in high winds? Like all tall, narrow buildings, the Tower sways a little. Here is the wind resistance of various parts

of the tower in winds of 192 km/h, with 320 km/h gusts: Antenna 6ft, 8in. from centre Sky Pod 3 ft, 4in. from centre

Tower Sphere 1 ft, 7in. from center (3)

2.7 Loading and strength of materials 2.7.1 Loading General All aspects of nominal loading (permanent, imposed, wind and other) at limit states (factored loads, Qd p = gD Dn + S(y igiQni), load reduction, etc. are covered in SANS 10160 (4) and the theory thereof in Chapter 6 of Structural Concrete Masonry A Design Guide (5). The following combinations of self-weight Dn, imposed floor loads Ln (or Qn latest revision in the Loading Code, (4)) and wind loads Wn, are generally considered at the ultimate limit state:

2.8

Antenna Sky pod Tower sphere

Figure 2.3: Members of the world federation of great towers

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1,5Dn 1,2Dn + 1,6Ln 1,2Dn + 0,5Ln + 1,3Wn 0,9Dn + 1,3Wn with the exceptions as given in the Loading Code. At the serviceability limit state the following combinations may be used: 1,1Dn + 1,0Ln 1,1Dn + 0,3Ln + 0,6Wn Note, the partial load factors, gI are combined to facilitate pattern loading on multiple spans of members to ensure the worst load combinations in determining the maximum bending moments and shear forces. If a simply supported single span member is considered then the maximum load combination is used and the maximum bending moment is found at mid-span. For the analysis of continuous beams (moments and shear forces), in accordance with SANS 10100 (2), clause 4.3.2.1 the following arrangements of loads should be considered: All spans loaded with total ultimate load (1,2Dn + 1,6Ln) (Figure 2.4a) ; All spans loaded with ultimate self-weight (1,2Dn ) and alternate spans loaded with ultimate imposed

load 1,6Qn. (Figure 2.4b). (a) (b) Figure 2.4: Pattern loading combinations 2.7.2 Material Ultimate Limit State The partial safety factor for concrete materials, gm are as follows: Concrete, gm = 1,5 in compression and in flexure (bending), and 1,4 in shear. Reinforcing steel, , gm = 1,15. Serviceability Limit State To calculate deflection, gm = 1,0 with respect to the reinforcing steel and the concrete. To calculate crack widths the values are respectively 1,0 and 1,3. This section of the work however will not be covered in this section.

1,2 Dn

1,6 Qn

1,2 Dn

1,6 Qn 1,6 Qn

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2.8 Analysis In the calculation of sectional properties of elements in a structure, the method and the model must emulate practice as close as possible. 2.8.1 Material properties Modulus of Elasticity of concrete, Ec (Youngs Modulus) Use the values for E as given in Table 1 in SANS 10100. In the case where special aggregate is used, such as light weight-aggregate the E value must be adjusted to cater for this material. Table 2.1: Values for Modulus of Elasticity of concrete, Ec.

1 2 Cube strength of concrete at the appropriate age or

stage under construction. MPa

Modulus of Elasticity of concrete, Ec GPa

20 25 30

40 50 60

25 26 28

31 34 36

Modulus of Elasticity of Steel, Es The modulus of elasticity of steel, (Es) is taken as 200 GPa. 2.8.2 Analysis of a section The analysis a cross-section of a reinforced concrete member is based on the short term strain relationship graphs given in Figure 2.5 and 2.6 for concrete and steel (Graphs 1 and 2 in SANS 10100).

Figure 2.5: Short-term design stress strain curve for normal density concrete.

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Figure 2.6: Short-term design stress strain curve for reinforcement. 2.9 Examples on loading 2.9.1 Example 1: Calculate the design load per running metre for an edge beam at ULS for a 250 suspended reinforced concrete floorslab in a housing complex.

SANS 10160 CALCULATIONS OUTPUT Nominal loads on edge beam:

Given App B-3.2 Table 4

Dn, beam = Dn, slab =

= = =

Ln, slab = =

35 kg/m = 35 x 10/1000 = 0,35 kN/m, with g = 10 m/s2 density of Cu x thickness x tributary width [kg/m3xmxm] 2 500 x 0,250 x 4, 5/2 kg/m (take Cu with nom As) 625 kg/m2x2,25m (alsoB-3.4 = mass/unit A=610 kg/m2) 1 406,25 kg/m = 1 406,25 x 10/1000 = 14,06 kN/m 1,5 kN/m2 ( Residential occupancy) x 4,5/2 m 3,38 kN/m

Dn, b = 0,35 kN/m Dn, s= 14,06 kN/m Ln, s= 3,38 kN/m

Ultimate load on edge beam: Design load effect Q 4.4.2

Q =

= = =

giD n= giDn +giQnj + S(yjgiQni) ij (1,5 SDn) = 1,5 x (0,35 + 14,06) = 21,62 kN/m, or. (1,2 Dn + 1,6 Qn) 1,2 x (0,35+14,06) + 1,6 x 3,38 = 22,70 kN/m

Q max 23 kN/m

2.9.2 Example 2: Rooftrusses for an inacessible roof are placed at 1,7 m centres and span 5 m. State the appropriate nominal imposed load Qn in kN/m2 acting on the roof.

SANS 10160 CALCULATIONS OUTPUT Nominal imposed load on load on an inaccessible roof: 5.4.4.3

Ln = Qn = =

(0,3 + (15 A)/60)kN/m2, & A =trib area = 1,7x5=8,5 m2 (0,3 + (15 8,5)/60) kN/m2 = 0,41 kN/m2

Ln 0,4 kN/m2

Discussion: With A =8,5 m2 & Ln = 0,4 kN/m2 the total Ln on area = 3,4 kN (8,5x0,4). Compare this with the concentrated load requirement of 0,9 kN; that will normally govern on short spans.

Nominal imposed load (Ln) for an accessible roof = 2,0 kN/m2 - Clause 5.4.4.2

250 reinforced concrete slab Dn beam = 35 kg/m 4 500

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2.9.3 Example 3: Calculate the windpressure (qz) on a building near Durban at the coast. The building is 25 x 7m in plan, 4 m in height and has a flat roof.

SANS 10160 CALCULATIONS OUTPUT 5.5 Nominal wind load, Wn: Fig 3 V = 40 m/s V = 40m/s Eq 5 (d) qz = kp Vz 2 velocity pressure 5.5.2.4 Terrain category 2: You have to design for either category 1 or 2

because the location is near Durban & therefore it cannot be category 3 or 4.

Terrain cat 2

5.5.2.6 Class B = Determination of forces on main structural member. Table 5 Kz = 0,92 (1,02 Cat 1) z = 4,0 m Fig 4 Kr = 1 (return period = 50 yrs) 5.5.2.1 a) Vz = V x Kr x Kz = 40 x 1 x 0,92 = 36,8 m/s(40,8m/s Cat 1) 5.5.2.1 c) > 24 m/s-OK for buildings with a 50-year return period 5.5.3.1 Kp = 0,6 (Sea level = 0 m) Eq 5 (d) Thus qz = kp Vz 2 = 0,6x(36,8)2 =813N/m2(999 N/m2) 0,8 kN/m2 qz = 0,8 kN/m

2 Now that the wind pressure has been calculated the aspect ratio of the building will determine the Cpe and

Cpis. The aforementioned calculation is basically followed on any type of structure, for example if the wind force on

a bulletin board has to be determined, the Cf value must be determined from the correct table taking the corresponding aspect ratios into account.

2.9.4 Example 4: Calculate Q max at ULS (Ultimate limit states) acting on a roof if Ln = 2 kN/m2 , Dn = 0,5 kN/m2 & Wn = + 0,3 kN/m2 (positive & negative pressure).

SANS 10160 CALCULATIONS OUTPUT Ultimate load on roof, Q:

4.4.2 Q1 = Q2 =

1,5Dn = 1,5 x 0,5 = 0,75 kN/m2, or 1,2Dn+1,6Ln = 1,2 x 0,5 + 1,6 x 2 = 3,8 kN/m2, or

Q3 =

= Q4 =

1,2Dn+0,5Ln+1,3Wn 1,2 x 0,5 + 0,5 x 2 + 1,3 x 0,3 = 1,99 kN/m2, or 1,3 Wn + 0,9 Dn = 1,3 x -0,3 + 0,9 x0,5 = 0,06 kN/m2

Q max= 3,8 kN/m2

Discussion: If Q4 was negative (stress reversal) then there would have been two design problems; i.e. the maximum loading Q2 and the second case for stress reversal.

2.9.5 Example 5: A 5 m tall column supports a tributary area of 35 m2 and is used as a filing space. Dn of the floor = 5 kN/m2 and Dn of the column is 45 kg/m. Calculate the load at ULS supported by the column.

SANS 10160 CALCULATIONS OUTPUT 5.4.3 b) Tributary A = 35 m2 > 20 m2 => Load reduction MAY be applied for

the area used as storage. Trib A = 35m2

Load red = 0,3 + 3,1/(A)1/2 = 0,3 + 3,1/(35)1/2 = 0,3 + 0,36 = 0,82 but with a minimum value of 0,5 => LR = 0,82

LR = 0,82

Table 4 (9)

Dn,floor = Dn, column =

Ln =

5 kN/m2 x 35 m2 = 175 kN for platform 0,45 x 5 = 2,25 kN for column 5 kN/m2 x 35 m2 = 175 kN for filing allow 5 kN/m2

4.4.2 Q = =

1,5Dn = 1,5(175 + 2,25) = 266 kN, or 1,2Dn+1,6LnxLR=1,2(175 + 2,25)+1,6 x175x0,82=442 kN

Q = 442 kN

2.9.6 Example 6: Calculate the overturning moment at ULS acting on a 2,0 m high masonry garden wall if Wn = 0,3 kN/m2 (Tip: the code also makes provision for an additional horizontal pointload).

Q

2.13

SANS 10160 CALCULATIONS OUTPUT Nominal loads on wall:

Given 5.4.5.3

Wn = 0,3 kN/m2, and also allow for a nominal horizontal concentrated force of 1,0 kN acting normal to the wall at any point at a height of 1,8 m & take worst case .

Wn = 0,3 kN/m2 or pointload

Ultimate bending moment on wall: 4.4.2

Mu = = = =

giWn x h2/2 1,3 x 0,3 x 22/2 = 0,78 kN-m/per m of walling, or giLn x h kN- m over length of whole wall 1,6 x 1,0 x 1,8 = 2,88 kN-m/m and take the nominal load as an imposed load.

Mu =0,78 kN-m/m, or 2,88 kN-m/m

Discussion: If the wall was 10 m in length, the lateral load and Mu is calculated per unity (1 m strip) and Mu still remains 2,88 kN-m/m due to the concentrated force applied at the top.

2.9.7 Example 7: The cross sectional area of a steel girder is 38 000 mm2. Calculate, Dn of the girder in kN per m.

SANS 10160 CALCULATIONS OUTPUT Nominal load of steel, Dn Take g steel = 7 800 kg/m3, thus

Appendix B-2.2

Dn = = =

area (m2) x density of steel (in kN/m3 ) (38 000 x (10-3)2 x 7 800 x 10/1000) 38 000 x 10-6 x 78 kN/m = 2,96 kN/m

Dn = 2,96 kN/m

2.9.8 Example 8: Calculate the bending moment at 1/4 span for a 10 m simply supported beam with a pointload in the centre. Q = 6kN/m en P = 10 kN (all loading is at ULS).

REFERENCE CALCULATIONS OUTPUT Reactions:

Rleft=Rright=

Mu span =

Q x l/2 + P/2 = 6 x 10/2 + 10/2 = 35 kN @ ULS Rleftx2,5Qx2,52/2=35x2,56x2,52/2=68,75 kN-m (left) Rrightx7,,5Qx7,52/2 P x 2,5 = 68,75 kN-m (right)

Mu 69 kN-m

2.9.9 Example 9: A round elevated steel water tank has a diameter of 8 m, is 2,5 m in height with a 10 mm wall thickness (Tank consist of base and side walls). The tank is supported on 4 columns, 2 m tall (Dn=60kg/m) and fixed at the bases. The roof of the tank is considered to be an accessible roof. Ignore other dead loads and calculate the ultimate vertical load per column and the moment on the structure due to the integrity load.

SANS 10160 CALCULATIONS OUTPUT Appendix B-2.2

Area tank =

Circ tank = Dn water =

Dn tank = Dn/column =

=

p x r2 = 50 m2 2 x p x r = 25,13 m = circumference of tank area tank x height x gwater = 50 x 2,5 x 10 = 1 257 kN (50 + 25,13 x 2,5) x 0,01 x g steel x 10/1000 = 88 kN (Dn water/4) + Dn column + Dn tank 1257/4 + 0,6 x 2 + 88/4 = 337,5 kN

A = 50 m2 Circ tank = 25,13m Dnwater=1257 kN Dn tank = 88 kN Dn col = 337,5 kN

5.5.4.2

Ln roof = Ln/column =

2,0 kN/m2 x 50 = 100 kN 0,93 kN 100/4 = 25 kN per column

4.4.2 Q/column = =

1,5Dn = 1,5 (337,5) = 506,25 kN, or 1,2 Dn+1,6 Ln= 1,2x 337,5 + 1,6 x 25 (ignore LR) = 445 kN

Q 506 kN

3.16 b 2) Integ Load = =

Moment =

=

1% x SDn (tank and columns)= (1/100) x (88 + 4 x 0,6 x 2) 0,93 kN integrity loading of self weight of structure

applied at mid-height of the tank 0,93 x (2x2,5/2) = 3,02 kN-m on structure (about base) 3,02/4 = 0,76 kN-m/column

Integload=0,93kN M=0,76 kN-m/col

Discussion: The integrity loading is a loading in lieu of the wind loading to ensure that the structural integrity of the structure is not compromised. This check must always be conducted by taking 1 % of the nominal dead load above the level under consideration.

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2.9.10 Example 10: Calculate the bending moment at quarter span for a 8 m simply supported beam with a pointload, P in the centre. Q = 6kN/m en P = 10 kN (loading is at ULS).

SANS 10160 CALCULATIONS OUTPUT Calculate reactions:

Ra = Rb = M span =

= =

10 + (6 x 8)/2 = 29 kN Ra x 2 Q x l2/2 29 x 2 6 x 22/2 46 kN-m

Ra = 29 kN Mspan=46 kN-m

2.9.11 Example 11: Calculate the loading at ULS for a 250 mm thick concrete residential floor slab. Give the answer in kN/m2.

SANS 10160 CALCULATIONS OUTPUT gcu = 24 kN/m3 Annexure B-3.2

T4 4.4.2

Dn slab = =

= Ln = Q = Q =

Depth x gcu [m x kN/m3] 0,25 x 24 [kN/m2] 6,0 kN/m2 1,5 kN/m2 (residential) 1,5 Dn = 1,5 x 6 = 9 kN/m2, or 1,2 Dn + 1,6 Ln = 1,2 x 6 + 1,6 x 1,5 = 9,6 kN/m2

Q = 9,6 kN/m2

Since the slab has not got a definite dimension like a beam, the loading can be calculated on a 1 m strip (unity) and the slab may be considered as a beam with a unit width. The one width (unity) is therefore similar to the next. If Q is multiplied by 1 m (taking that as the width of the slab), Q = 9,6 kN/m2 x 1 m can therefore be expressed as Q = 9,6 kN/m. This loading is therefore applied on each strip = to unity.

2.9.12 Example 12: What is the minimum loading for movable partition walls on a slab ?

SANS 10160 CALCULATIONS OUTPUT 5.4.1.4 Dn = 1 kN/m2 for partitions having a weight less 3 kN/m Dn = 1 kN/m2

2.9.14 Example 14: Determine the nominal horizontal loading, expressed as kN/m to be applied on a 1,1 m high balustrade wall with a length of 10 m.

SANS 10160 CALCULATIONS OUTPUT 5.4.4.1 Ln = 0,5 kN/m balustrade applied at top of wall Ln = 0,5 kN/m

2.9.15 Example 15: A 230 thick brick wall is 2,7 m in height and 6 m in length and supports a vertical 20 kN/m imposed load at the top. Calculate the maximum load at ULS supported at the base of the wall.

SANS 10160 CALCULATIONS OUTPUT gbrick = 260 kN/m2 for 120 mm wide wall = 260 x 230/120 x 10/1000 = 4,9 kN/m2 of wall.

Annexure B-5.1 Given 4.4.2

Dn wall= Ln =

Q1 = Q2 =

height x gbrick = 2,7 x 4,9 = 13,46 kN/m of wall 20 kN/m 1,5 Dn = 1,5 x 13,46 kN/m = 20,18 kN/m, or 1,2 Dn + 1,6 Ln = 1,2 x 13,46 + 1,6 x 20 = 48,15 kN/m

Dn wall=13,46 kN/m Ln = 20 kN/m Q = 48 kN/m

2.9.16 Example 16: Three reinforced concrete beams (Dn = 45 kg/m) are placed at 2,2 m centres and span 9,5 m. The beams support a 275 thick reinforced concrete slab. Calculate Q max per m, at ULS for the central supporting beam if Ln = 3 kN/m2.

BM diagram Bm span = 46 kN-m

2.15

SANS 10160 CALCULATIONS OUTPUT Given B3-2 T4

Dn beam = Dn slab =

Ln = =

45 kg/m = 0,45 kN/m thickness x width x gcu = 0,25 x 2,2 x 24 = 14,52 kN/m 3 kN/m2 (design for Ln = 3 kN/m2) 3 x 2,2 = 6,6 kN/m.

Dn b= 0,45 kN/m Dn s = 14,52 kN/m Ln = 6,6 kN/m

4.4.2 Q = =

Q =

1,5 Dn = 1,5 x (0,45 + 14,52) 22,5 kN/m, or 1,2 Dn + 1,6 Ln = 1,2.(14,97) + 1,6 x 6,6 28,5 kN/m

Q = 28,5 kN/m

2.9.17 Example 17: Rooftrusses for an inacessible roof are placed at 6 m centres and span 8 m. State the appropriate imposed load Ln in kN/m2 acting on the roof.

SANS 10160 CALCULATIONS OUTPUT 5.4.3.3 b Ln = (0,3 + (15 A)/60) with

A = tributary area = 6 x 8 = 48 m2 > 15 m2 Nominal imposed load (Ln) = 0,3 kN/m2

Ln = 0,3 kN/m2

2.9.18 Example 18: Calculate the maximum moment at ULS that a communication mast, 17 m tall must be designed for if Wn = 0,5 kN/m.

SANS 10160 CALCULATIONS OUTPUT Table 2 Mu =

= =

1,5 x Wn x h2/2 (use gi = 1,5 for freestanding, or 1,3) 1,5 x 0,5 x 172/2 109 kN-m at base of mast.

BM = 109 kN-m

2.9.19 Example 19: A rectangular reinforced concrete beam is 650 mm deep and 250 mm wide. The beam is simply supported over a span of 8 m and carries a uniformly distributed (live) load of 10 kN/m. The beam also carries a point load of 40 kN (dead load) at 2 m from the left support. a. Draw the load diagram for the design of the beam. Indicate values on the beam diagram. b. Draw a shear force and bending moment diagrams for the design of the beam. Indicate

the position and value of the maximum bending moment. REFERENCE CALCULATIONS OUTPUT

SANS 10160 Appendix B-3.4

Nominal self-weight Concrete beam (Dn) = 24kN/m3 x 0,65 x 0,25 = 3,9 kN/m

Dn,beam=3,9kN/m

Ultimate load on beam, Q: 4.4.2 5.4.1.3b)

Q = i) Q =

= or, ii) Q =

= =

=

(1,5 Dn) 1,5 x 3,9 kN/m = 5,85 kN/m (udl), & 1,5 x 40 = 60 kN Pointload (1,2 Dn) + (1,6 Ln) 1,2 x 3,9 + 1,6 x 10 20,68 kN/m (udl) & 1,2 x 40 = 48 kN Pointload

Q = 5,85 kN/m & = 60kN Pointl Q = 20,68 kN/m & = 48kN Pointl

Line diagram Case 1

60 kN & udl = 5,85 kN/m 2 000 6 000 Line diagram 68,4 56,7 + 3,3 - SF [kN] 38,4 kN + BM [kN-m] 125,1 (max)

2.16

REFERENCE CALCULATIONS OUTPUT

For case 2: Ultimate load on beam, Q:

Q = =

For Ra: Ra =

= For Rb:

Rb = =

20,68 kN/m (udl) & 48 kN Pointload 3MB = 0 (20,68 x 8 x 8/2 + 48 x 6)/8 118,72 kN 3MA = 0 (20,68 x 8 x 8/2 + 48 x 2)/8 94,72 kN

Ra = 118,72 kN Rb = 94,72 kN

Checks: 3V = 0 Ra + Rb = 118,72 + 94,72 = 213,44 kN

Udl + Pointload = 20,68 x 8 + 48 = 213,44 kN => OK

Max BM x =

BM max = 94,72/20,68 = 4,58 m 4,58x94,7220,68x4,582/2 @ 4,58 m from right support

BMmax =217 kN-m

REFERENCE CALCULATIONS OUTPUT

Line diagram Case 2

2.9.20 Example 20: A four storey braced residential building with a roof (all in reinforced concrete 300 mm thick flat slab construction, without any beams) has a 4 x 4 m grid layout. All columns between floors and floor and roof and slab and footings have a clear height of 3 000. Specify a column with dimensions to ensure a short column dimension. No provision is made for the self weight of walls on the slab. Calculate the bearing pressure under a base of a typical centre column. You need not apply live load reduction.

48 kN & udl = 20,68 kN/m

2 000 6 000 Line diagram 118,72 + 77,36 29,36 - SF [kN] 1 420 x =4 580 94,72 + BM [kN-m] 217 (max)

2.17

SANS 10160 CALCULATIONS OUTPUT

5.4.4.2 5.4.3.2 5.4 Table 4 .1

Table 2

Select a column that will be short or stubby with effective length of

2700. Select a 300 x 300 column. To calculate the ground bearing pressure, the axial load, N has to

be calculated at SLS conditions. ULS load factors are applied to design the reinforced concrete base. The central column is symmetrical and supports a 4 x 4 m tributary area of all slabs and therefore a concentric axial load , N.

Dn floor slabs = 3 floor slabs & 1 roof slab (all similar flat slabs) = 4 x (4 x 4 x 0,3 x 24 kN/m3) = 460,8 kN

Dn columns = total column length x cross section x g concrete = (2,7 x 4) x (0,4 x 0,4) x 24 kN/m3 = 41,47 kN

Dn base say = (2,5 x 2,5 x 0,6) x 24 kN/m3 = 90 kN 3Dn = 461 + 42 + 90 = 593 kN Ln roof slab = tributary roof area x 2 kN/m2 (access roof)

= 4 x 4 x 2 = 32 kN Ln residential = 3of all tributary floor areas x 1,5 kN/m2

= 3 x (4 x 4) x 1,5 = 72 kN 3Ln = 32 + 72 = 104 kN N = gI x (3Dn + 3Ln) = 1,0 x (593+104) = 697 kN

Earth pressure under base = 697/area = 697/(2,5 x 2,5) = 112 kN/m2 Note: If the earth pressure was calculated at ULS conditions you would have been out by 50% and a disaster as a technologist.

Specify a 400 x 400 column Dn slabs = 461kN Dn columns= 42kN Dn base = 90kN 3Dn = 593 kN

3Ln = 104 kN N = 697 kN s = 112 kN/m2

2.9.21 Example 21: A 600 deep beam x 300 wide is continuous over three x 6 000 spans. The imposed load on the beam is 5 kN/m. Use Table 4 in SANS 10100 and calculate and draw the bending moment diagrams for the beam.

SANS 10160 CALCULATIONS OUTPUT Appendix B Given 0100 cl 4.3.2.2

Nominal self-weight (Dn) = 2 400x0,6x0,3x10/1000 = 4,32 kN/m Nominal imposed load (Ln) =5,0 kN/m Qn/Dn = 5,0/4,32 = 1,16 < 1,25 => Table 4 may be used

Dn = 4,32 kN/m Ln = 5,0 kN/m

Ultimate load on slab, Q: SANS 10160 4.4.2

i) Q = or, ii) Q =

=

(1,5 Dn) = 1,5 x 4,32 = 6,48 kN/m (1,2 Dn) + (1,6 Ln) 1,2 x 4,32 + 1,6 x 5,0 = 13,18 kN/m

Q = 13,18 kN/m

Bending Moment: SANS 10100 T4 Near middle end span First interior support

Middle of interior span

= QL/11 = 13,18 x 6 x 6/11 = 43,14 kN-m = -QL/9 = 43,14 x 11/9 = -52,72 kN-m = QL/14 = 43,14 x 11/14 = 33,90 kN-m

4 000 4 000

Roof Floor 3 Floor 2 Floor 1 Central column @ 4 x 4 matrix Floor 0 on ground

2.18

SANS 10160 CALCULATIONS OUTPUT Shear Force: SANS 10100 T4

At outer support First interior support

At interior support

= 0,45 Q = 0,45 x 13,18 x 6 = 35,59 kN = 0,6 Q = 35,59 x 0,6/0,45 = 47,45 kN = 0,55 Q = 35,59 x 0,55/0,45 = 43,5 kN

2.9.22 Example 22: A 4 000 x 5 000 reinforced concrete slab is 170 thick and fully supported on 230 masonry load bearing walls. A 100 concrete masonry wall is 2 700 in height is place in the middle over the short and another parallel 1 000 to the left and another parallel a 1 000 to the right, three times as indicated. Calculate the total nominal dead load the slab has to be designed for.

SANS 10100 CALCULATIONS OUTPUT

SANS 10160 Appendix B-3.2 B-3.2 B-5.3

Nominal self-weight slab (Dns) = 2 400 x 0,17 x 10/1000 = 4,0 kN/m2 Allow for 50 mm finishes on the slab, take gfinishes = 24 kN/m3 Nominal self-weight fin (Dnf) = 2 400 x 0,05 x 10/1000 = 1,2 kN/m2 Nominal self-weight solid concrete bricks (Dn) =440/0,2 kg/ m2

= 2 200 kg/m2 for concrete walling say Total wall (Dn) = 2 200 x 0,10 x2,7 x 3,540 x 10/1000 = 21 kN Nominal self-weight walls (Dn)/ m2 = 21x3/(4,54x 3,54) =3,93kN/m2 3Dn = 4,0 + 1,2 + 3,93 =9,13 kN/m2

Dns = 4,0 kN/m2 Dnf = 1,2 kN/m2 Walls: Dn = 1,31kN/m2 3Dn =3,93kN/m2 All: 3Dn = 9,13 kN/m2

If a single wall was placed in the centre of the slab as shown spanning in the 4 000 direction, a slab of equal to unity (1 m) could have been strengthened and designed to support the wall; the same can be done for the three walls spaced at 1 000 centres.

However, another approach is to take the self-weight of the inner walls on a slab and to treat it as a uniformly distributed load over the slab area.

If the wall had spanned in the long direction, i.e. 5 000, the slabs = unity can be designed treating the wall as a point load.

43,14 33,90 6 000 6 000

-52,72 BM diagram [kN-m]

33,59 + 43,5 +

6 000 6 000

- 47,45 -43,5

SF diagram [kN] Centreline

2 500 2 500

Wall in middle 230 wall 4 000

2.19

2.10 References 1. South African Bureau of Standards. The general procedures and loadings to be adopted in the

design of buildings. SANS 10160: 1989. 2. South African Bureau of Standards. Code of practice, The structural use of concrete, Part 1:

Design. SANS 10100-1:1992. 3. South African Bureau of Standards. Code of practice, The structural use of concrete, Part 2:

Materials and execution of work. SANS 10100-2:1992. 2.11 Assignments 2.11.1 Assignment 1 Revise all previous work using the equilibrium equations to calculate loads, shear forces, bending moments and the location of maximum bending moments for static determinate structures. 2.11.2 Assignment 2 Calculate the maximum bending moment and shear forces for the beam with the cantilever section (Figure 2.7). Dn = 10 kN/m and Ln = 12 kN/m. Consider the following load cases. (i) Maximum load on AB and BC. (il) Maximum load on AB, minimum on BC. (iii) Maximum load on BC, minimum on AB. Figure 2.7: Beam with cantilever 2.11.3 Assignment 3 Work through the following Compact programmes: The programme is available in the computer laboratory or can be downloaded at the following web sites: http://www.shef.ac.uk/~compact, or : http://www.cnci.co.za Conceptual design, section 1 - 7: Reinforced concrete Design Part 1 up to section 6 Loads When you can show progress of at least 70% to the laboratory assistant for each section, you will

be considered competent in this section; the lab assistant will indicate this mark on a class list. List all obvious code differences between BS 8110 and SANS 10100 and submit on one A4 sheet

to the laboratory assistant.

A B C 4 000 1 500