Beams design and analysis

3.0 Design of Beams3.1 Rectangular Beam3.1.1 Singly Reinforced Beam

Example 3.1

Uniformly distributed characteristic load (udl)Dead = 3.76 kN/m (excluding beam self weight)Imposed = 10kN/m

Design the beam if fy = 460N/mm2, fcu = 30N/mm2 and density of concrete ρconcrete = 24kN/m3

Soln(1) Loading

Beam self-weightDead load Imposed loadTotal design load

(2) Bending

Bending moment ,l = effective span

Effective span (l) = Clear span + half support width at each end or Clear span + half effect depth at each end whichever is less (BS 8110 Part1: 1985 clause 3.4.1.1)

Using formulae, using table for z/d or using a design chart we can now calculate the area of steel.(a) Formulae

From the steel area table app pg 7

1

A

A

A-A

450

1006900100

300

400

4T20 – B (4no high yield type two bars in the bottom of the beam)(1260mm2)

Note: 106 changes moment from kNm to Nmm, in practice there has been tendency to use Y in place of T and vice versa, however it should be noted that this is a mistake b’se Y is not identical to T. T has higher bond strength than Y Y stands for high yield deformed steel type one (square twisted bar) nowadays is obsolete (outdated)T stands for high yield deformed steel type two (Ribbed bar) nowadays is extensively usedR stands for round mild steel.

(b) TablesAs above k = 0.11z/d = 0.857 from table, Appendix page 7

4T20 – B(1260mm2)

(iii) Chart

, From chart 2, appendix page 5

4T20 – B(1260mm2)

The minimum and maximum areas of tension steel in a beam are given in clause 3.12.6.1 and table 3.27 of BS8110 see appendix pg 8Min allowable = 0.13%Max allowable = 4% of the concrete section (bxh)

Actual % As

0.13<0.93<4% therefore the max and min is ok

3.1.1.2 Design in shearThe actual behaviuor of reinforced concrete under shear is very complex and still the subject of debate among the researchers. However design techniques have been developed which work in practice. To determine the shear force supported by the link reinforcement, consider the free body diagram to one side of the crack shown in fig 3.1. Clearly only bars crossing the crack contribute.

2sv

d

Fig 3.1 the free body to one side of an assumed 45˚ shear crack.

45˚

45˚

Number of bars

Shear force supported by shear links/stirrups (Vs)

Asv = Area of all legs of each shear link Sv = spacing of linksfyv = characteristic strength of shear reinforcement

Converting the above force to an equivalent stress over the whole beam, vs, by dividing by the cross sectional area of the concrete, bd

For the beam to remain safe; the applied shear stress should be equal to the shear capacities contributed by the steel and concrete. Thus summing the contributions from the steel and concrete we have:-

In minor beams a minimum area of links that can support a shear stress of 0.4N/mm2 must be provided throughout the whole beam.

Minimum

Note: In practice, in the design of beam, after determining the required reinforcement for bending it is then necessary to check the beam for shear.First the shear forces (V) are calculated at the critical section. Then an average shear stress is calculated by dividing the shear force (V) by the effective cross sectional area (bxd) of the beam (clause 3.4.5.2). This stress is compared with the shear capacity of concrete and any excess stress is catered for by providing shear links. Table 3.8 gives the requirement for shear reinforcement and table 3.9 gives the shear stress that the concrete can carry.

3V2

V1

Shear Force Diagram

dV

Shear crack at approx. 45˚

Fig 3.2Shear stress must be checked as follows:-

Check 1 or whichever is less. This check is sometimes referred to as check for

section dimensions for shear (clause 3.4.5.2).

Check 2 if , then nominal stirrups (links) of area

Or if , then designed stirrups (links) of area

Refer example 3.1(3) ShearFrom fig 3.2

Shear force

The distance from V to the mid span is

The distance from V1 to the mid span is The distance from V2 to the mid span is

By similar triangles

Shear stress

Section dimensions are ok

Using table 3.9 for finding shear capacity of the concrete, the % area of tension steel effective at the beam end must be known. In example 3.1 only two tension bars extend to the support. Thus area = 628mm2 for 2T20

Then we need to interpolate in table 3.9 btn 0.5 and0.75%, the column for

4

% at 0.50 vc = 0.50 at 0.75 vc = 0.57

at 0.523 vc =

Note: Table 3.9 applies to a C25 concrete (see note2)

Thus the actual vc for C30 (vc30)

Nominal stirrups (links) of area

But the maximum spacing of link (svmax) is 0.75d (see clause 3.4.5.5 of BS 8110)

Note: each stirrup passes one side of the beam and back down the other. Its area thus counts twice.By reference to steel area table on pg 7 of the appendix the likely areas are:-2R8 = 101mm2, 2R10 = 157mm2, 2R12 = 226mm2. Here R10 will be used because R8 will give more stirrups while R12 will give the spacing greater than the recommended. If R10 are used will have to be at closer center than 300mm.

Thus maximum centers

It is normal to space stirrups (links) at centers of multiples of 25, the spacing will be 275mmR10-275

2Nominal (not designed) bars are to be added at the beam top to support stirrups (links). Their diameter is normally half diameter of the tension steel but in the order of 20,16 and 12.

Total number of stirrups (links).

3.1.1.3 Design in deflectionAfter determining load, bending and shear finally we have to check for the beam in deflection. In monolithic concrete frames actual deflection is not calculated b’se is inaccurate and time consuming. Instead we have to compare the actual span /effective depth ratio of the beam with the allowable span /effective depth ratio defined in table3.10, 3.11 and 3.12(for doubly reinforced beam)

Refer example 3.1

Actual span /effective depth ratio

Allowable span /effective depth ratio = Factor from table3.10 x Factor from table 3.11

From table 3.10 app. pg1 basic span/effective depth ratio for simply supported rectangular beam = 20

Modification factor for tension reinforcement from table 3.11 app. pg 4 given by note 1 is:-

Modification factor

5

From note 2 app pg 4 for simple supported system βb = 1

Modif.

Allowable span /effective depth ratio

Allowable span /effective depth ratio =19.52 > Actual span /effective depth ratio =17.5,thus deflection is Ok

Note: After completing the design, the RC detail and bar bending schedule should be prepared (Referred to as complete structural working drawings). ); however this is discussed in 3.3

3.1.2 Doubly reinforced beam

6

When M>0.156bd2fcu the design ultimate moment M exceeds the moment of resistance of the concrete, the compression reinforcement is required. For this condition the depth of neutral axis, , the maximum value allowed by the code in order to insure a tension failure with a ductile section

Therefore eqn 5 becomes

Cc = Compression force in concreteCs = Compression force in steelTs = Tensile force in steelAs’ = Area of Compression steelAs = Area of tension steel

For equilibrium of the section aboveTs =Cc+Cs

with

Taking the moment about the tension steel As

But

Multiplying both sides of eqn 11 by z = 0.775d

7

Z=lever arm

Cs

0.67fcu/ɣm

d’

b

Beam section

stress block

x =

0.5d

0.9x

dN/A

Fig 3.3 beam section with compression reinforcement

As’

Ts

Cc

As

Example 3.2 Doubly reinforced

A simply supported beam of effective span 7000mm is 500mm deep by 260mm wide. It is made of grade 30 concrete and is required to give 2hrs fire resistance in mild exposure. The beam carries a characteristic dead load of 12.9kN/m in addition to its self-weight and characteristic imposed load of 15kN/m.The beam frames in to 300mm at each end.If fy = 460N/mm2 design the beam

(1) LoadingSelf wt of the beamDead load Imposed load

Total design load

(2)Bending

Bending moment

Cover: exposure 25mm table 3.4 Fire 40mm table 3.5Effective depth d = h- (cover + dia of link (stirrup) +0.5dia of tension steel)Assume T20 and R10 will be used as tension and shear link resp.

Effective depth

Doubly reinforced

d’= cover +stirrup dia +0.5dia of compression barAssume compression bar to be T20

From steel area table app pg 72T32+1T25 – B(2101mm2)

2T16-T(402mm2)Check for maximum and minimum allowable

8

The minimum and maximum areas of tension and compression steel in a beam are given in table 3.27 (appendix pg 8) and clause 3.12.6.1 of BS8110

(a) Tension steelMin allowable As = 0.13%Max allowable As = 4% of the concrete section (bxh)

Actual % As


(b) Compression steelMin allowable As’ = 0.20%Max allowable As’ = 4% of the concrete section (bxh)

Actual % As’

0.20 <0.31<4% therefore the max and min is ok

(3) Shear

From fig above

Shear force

The distance from V to V1 is


By simple analysis (refer to CET 040208 structural mechanics in NTA 4)

Upward positive

9

V V10.15m

46.43kN/m

V V20.59m

46.43kN/m

V2V1

Shear Force Diagram

dV


Upward positive

Shear stress


This must be compared with vc

Using table 3.9 for finding shear capacity of the concrete, the % area of tension steel effective at the beam end must be known. In this example only two tension bars extend to the support. Thus area = 1610mm2 for 2T32

Then we need to interpolate in table 3.9 btn 1.0 %and1.5%, the column for

% at 1.0 vc = 0.63 at 1.5 vc = 0.72

at 1.41 vc =



designed stirrups (links) of area

But the maximum spacing of link (svmax) is 0.75d (see clause 3.4.5.5 of BS 8110)

By reference to steel area table on pg 7 of the appendix the likely areas are:-2R8 = 101mm2, 2R10 = 157mm2, 2R12 = 226mm2. Here R10 will be used because R8 will give more stirrups while R12 will give the spacing greater than the recommended.

Thus maximum centers



10

(4) Deflection


Allowable span /effective depth ratio = Factor from table3.10 x Factor from table 3.11x Factor from table3.12

From table 3.10 app. pg1 basic span/effective depth ratio for simply supported rectangular beam = 20


Modification factor


Modif.

Modification factor for compression reinforcement from table 3.12 app. pg 1 can be determined after

calculating the value of

Factor = 1.1



Note: After completing the design, the RC detail and bar bending schedule should be prepared (Referred to as complete structural working drawings) ); however this is discussed in 3.3

3.2 Flanged beamsIn many monolithic RC structures the beam can make a use of a part of the slab to act as a compression flange. This is only true for span (sagging) moments. At the support hogging moment will place the bottom the beam in compression.

11

flange width bf

flange thickness hf

Webwidth bw

For T beam shown aboveFlange width bf = web width (bw) +lz x 0.2 or the actual width whichever is less (clause 3.4.1.5(a))

For L beam shown aboveFlange width bf = web width (bw) +lz x 0.1 or the actual width whichever is less (clause 3.4.1.5(b))

Where lz is the distance btn points of zero moment (which, for a continuous beam, may be taken as 0.7 times the effect span)

For T-sections and L sections which have their flanges in compression can both be designed or analysed in a similar manner, and the eqns which are derived can be applied to either type of cross section. As the flanges generally provide a large compression area, it is usually unnecessary to consider the case where compression steel is required.For the singly reinforced section it is necessary to consider two separate cases:-

(a) The stress block lies within the compression flange(b) The stress block extends below the flange

3.2.1 Flanged section with the depth of the stress block lies within the flange 0.9x < hf

The fig above shows the T section, stress block within the flange 0.9x < hf

For this depth of stress lock, the beam can be considered as an equivalent rectangular section of breadth bf equal to the flange width. This is because the non- rectangular section below the neutral axis is in tension and is, therefore considered to be cracked and in active.

Therefore

12

flange width bf

flange thickness hf

Webwidth bw

TS

CC

Section

Stress

Z

0.67/ɣm

0.9x

N/A d

bf

hf

bw

3.2.2 Flanged section with the depth of the stress block extending below the flange 0.9x > hf

An alternative procedure to check if he depth of the stress block extends below the flange, is to calculate the moment of resistance, Mf, of the flange (section) with 0.9x = hf, the depth of the flange.

Moment of resistance, Mf, of the flange is

If the design moment Md > Mf, then the stress block must extend below the flange and 0.9x > hf

In this case the design can be carried out by either (a) Using an exact method to determine the depth of the neutral axis or(b) Designing for the conservative condition of x = d/2.

Fig below shows Flanged section with depth of neutral axis x = 0.5d

Calculating areas 1 and 2 shown in the fig aboveArea 1 Area 2 With x = 0.5dArea 2 Compression forces developed by these areas are:-

Taking moments about Cc1 at the centroid of the flange

This eqn is given in pg 30 of book and as eqn 1 in clause 3.4.4.5 of BS 8110Note: This eqn should not be used when hf > 0.45

Example 3.3 Flanged beam

13

Z1

2CC2

TS

CC1

SectionStress

Z2

0.67/ɣm

0.9x

N/A d

bf

hf

bw

1 1

A simply supported T beam with an effective length of 6000mm is made of grade 30 concrete and is required to give 2hrs fire resistance in mild exposure. The beam carries a characteristic dead load of 30kN/m (excluding its self wt) and characteristic imposed load of 9kN/m. The beam frames in to 300mm columns at each end. Given fy = 460N/mm2 and geometrical properties of the section shown below, design the beam

(1) Loadingbf = bw + 0.2lz or the given value whichever is less

or 400mmbf = 400mm

Self-wtDead loadLive load

Ultimate load

(2) Bending

Bending moment

Cover: exposure 25mm table 3.4 Fire 40mm table 3.5Effective depth d = h- (cover + dia of link (stirrup) +0.5dia of tension steel)Assume T20 and R10 will be used as tension and shear link resp.

Effective depth

z/d = 0.903(from Lever Arm Tablez = 0.903dFrom eqn 5

0.9x = 105<hf = 120 the stress block lies within the flange.Alternatively we can check if the stress block lies within the flange by comparing the moment of resistance Mf of the flange and the design moment Md. IF Mf > Md, then stress block lies within the flange, otherwise the stress block extends below the flange

Mf = 309>Md=279 the stress block lies within the flange.

14

600mm

400mm

120mm

250mm

From steel area table app pg 73T25 – B(1470mm2)

Check for maximum and minimum allowable

The minimum and maximum areas of tension and compression steel in a beam are given in table 3.27 (appendix pg 8) and clause 3.12.6.1 of BS8110

Therefore max and min As are as indicated below

Min allowable As = 0.13%Max allowable As = 4% of the concrete section (bxh)

Actual % As


(3) Shear

From fig above

Shear force



By simple analysis (refer to CET 040208 structural mechanics in NTA 4)

Upward positive

15

V V10.15m

62kN/m

V2V1

Shear Force Diagram

dV


Upward positive

Shear stress


This must be compared with vc

Using table 3.9 for finding shear capacity of the concrete, the % area of tension steel effective at the beam end must be known. In this example only two tension bars extend to the support. Thus area = 982mm2 for 2T25

Then we need to interpolate in table 3.9 btn 0.5 and 0.75% the column for

% at 0.5 vc = 0.5 at 0.75 vc = 0.57

at 1.41 vc =



Designed stirrups (links) of area

The maximum spacing of link (svmax) is 0.75d (see clause 3.4.5.5 of BS 8110)


2Nominal (not designed) bars are to be added at the beam top to support stirrups (links). Their diameter is normally half diameter of the tension steel but in the order of 20,16 and 12.


16

V V20.69m

62kN/m

(4) Deflection


Allowable span /effective depth ratio = Factor from table3.10 x Factor from table 3.11x Factor from table3.12

From table 3.10 app. pg1 basic span/effective depth ratio for simply supported flanged beam

thus linear interpolation should be applied (clause 3.4.6.3)

20 a 16


Modification factor


Modif.



Note: After completing the design, the RC detail and bar bending schedule should be prepared (Referred to as complete structural working drawings); however this is discussed in 3.3

17

3.3 Preparation of structural drawings.3.3.1 Introduction To be able to prepare detailed structural drawings we need to know the following first:-

3.3.2 Anchorage length The reinforcement bar subjected to tension must be properly anchored otherwise may be pulled out of the concrete.

The anchorage length depends on the bond between the bar and the concrete, and the area of contact.

Let, L = be the minimum anchorage length to prevent pull outF = the nominal diameterfbu = ultimate anchorage bond stressfs = direct compressive/tensile stress in the barF = applied tensile force

18

L

anchorage bond stress fbu

FF

Consider the forces in the bar Tensile pull out force F = Bar’s cross- sectional area × direct stress

Anchorage force = Contact area × anchorage bond stress In order this system to remain in equilibrium these two forces should be equal

Anchorage length L =

Anchorage length L =

The design ultimate anchorage bond stress, fbu is given by the equation

Therefore the Anchorage L =

bis the bond coefficient which depends on bar type and the type of force the bar is subjected to.

The values of b are given in table 3.28 of BS 8110.Table 3.28 Values of bond coefficient bBar type Bar in tension Bar in compressionPlain bars 0.28 0.35Type1:deformed bar (Y) 0.40 0.50Type2:deformed bar (T) 0.50 0.63Fabric 0.65 0.81(Note: Type 2 (T) ribbed, Type 1(Y) square twisted inferior bond characteristic)

Example 3.4Determine the length of tension anchorage required for the 25mm diameter plain mild steel reinforcement (R25) in the cantilever beam made of grade 30 concrete

19

Anchorage Length L

Effective span

Solution

Say 910mm

3.3.2.1 Anchorage of bars at simply supported end of a member (clause 3.12.9.4)

At simply supported end of a member, each tension bar should be anchored by one of the following

(a) An effective anchorage length equivalent to 12 times the bar size beyond the centre line of the support; no bend or hook should begin before the centre of the support.

(b) An effective anchorage length equivalent to 12 times the bar size plus d/2 from the face of the support, where d is the effective depth of member; no bend or hook should begin before d/2 from the face of the of the support.

(c) For slabs, if the design ultimate shear stress at the face of the support is less than the appropriate value, vc, recommended in (clause 3.4.5), a straight length of bar beyond the centre line of the support equal to either one third of the support width or 30mm, whichever is the greater.

3.3.2.2 Anchorage of links (clause 3.12.8.6)A link may be considered to be fully anchored if it satisfies the following:

(a) It passes around another bar of at least its own size, through an angle of 90°, and continues beyond for a minimum length of eight(8) times its own size; or

(b) It passes around another bar of at least its own size, through an angle of 180°, and continues beyond for a minimum length of four(4) times its own size;

3.3.3 LapsLapping of reinforcement is often necessary to transfer the forces from one bar to another. They should be placed, if possible, away from points of high stress and should preferably be staggered (clause 3.12.8.9)

3.3.3.1 Minimum laps (clause 3.12.8.11)

The minimum lap length for bar reinforcement should be not less than 15 times the bar size or 300mm, whichever is greater, and for fabric reinforcement should be not less than 250mm.

3.3.3.2 Tension laps (clause 3.12.8.13)

The tension lap length should be at least equal to the design tension anchorage length (see 2.2 above) necessary to develop the required stress in the reinforcement. Lap length for unequal size bars (or wires in fabric) may be based upon the smaller bar. The following provisions also apply:

20

(a) Where a lap occurs at the top of a section as cast and the minimum cover to either is less than twice the size of the lapped reinforcement, the lap length should be increased by a factor of 1.4;

(b) Where a lap occurs at the the corner of section and the minimum cover to either face is less than twice the size of the lapped reinforcement or, whwre the clear distance between adjacent laps is less than 75mm or six times the size of the lapped reinforcement, whichever is the greater, lap length should be increased by a factor of 1.4;

(c) In cases where both conditions (a) and (b) apply, the lap length should be increased by a factor of 2;

3.3.3.3 Compression laps (clause 3.12.8.15)

The compression lap length should be at least 25% greater than compression anchorage length (see 3.3.3.2 above) necessary to develop the required stress in the reinforcement. Lap length for unequal size bars (or wires in fabric) may be based upon the smaller bar.

3.3.4 Curtailment of reinforcementCurtailment of reinforcement in reinforced concrete structural working drawing in accordance with the standards is necessary in order to avoid providing redundant reinforcement.

3.3.4.1 Curtailment of reinforcement in beams (clause 3.12.10.2)The simplified curtailment rules illustrated in figure 3.24 on app pg 9 of book1 design tables to BS 8110 may be used for beams in the following circumstances.

(a) The beams are designed for predominantly uniformly distributed loads.

(b) In the case of continuous beams, the spans are approximately equal.

After being conversant with literature above; the complete structural drawings (ga, rc details and bar bending schedules) are prepared as we shall see/discuss in the class.

21

Tutorials BEAMSQ1For the cantilever “T” – beam of 4m length find the maximum design load w, if the reinforcement consist of 2 high yield steel bars of 20mm dia. (As = 6.28 cm2), fy = 460N/mm2 and the concrete is grade 30, fcu = 30N/mm2. The beam has a cross section as shown in Fig. 1

w=? 400mm

100 4000

d =500

FIG1 200 Q2An overhanging reinforced concrete beam carries a uniformly distributed load of 5kN/m and concentrated loads of 20kN at the ends of the beam (Fig.2). 20kN 20kN

5kN/m

22

2m A 6m B 2m

FIG 2

(i) Determine the support reactions and the maximum bending moment. Draw the bending moment diagram.

(ii) Sketch the pattern of main reinforcement for the beam. (iii) Using a rectangular stress block define the depth of compressive zone x and the ultimate

moment of resistance Mu if the effective depth of the beam 400mm, width of section 200mm, cross – sectional area of bars 628mm2. Given the characteristics strengths are fy = 425N/mm2 for the reinforcement, and fcu = 25N/mm2 for the concrete. Check the resistance of the beam.

Q3 A simply supported T-beam of 6m length carries a uniformly distributed load of 30kN/m (including self weight). The main geometrical characteristics of the cross-section are as shown in figure1. The materials to be used are grade 30 concrete, fcu =30N/mm2 and high yield steel of strength fy =425N/mm2. Determine the area of steel required at mid-span, assuming the effective depth of the beam is 0.9h.

23

FIG.1 ALL DIMENSIONS ARE mm

Q4

By mistake a contractor provided 2Y32 and 1Y25 tension steel to a rectangular beam. His poor workmanship produced concrete of strength 28N/mm2 while the dimensions of the beam section were as per design which is b/d = 300/700mm. If the characteristic strength of the steel is fy = 410/mm2,use a simplified rectangular stress block to calculate the maximum bending moment which the beam can safely support.

Q5Shown in the figure below is a rectangular section of a beam and a rectangular stress block distribution. Obtain the expression for the maximum ultimate resistance for the section.

Q6a) A rectangular beam b/d = 300mm/700mm is to resist a bending moment of 625kNm. Using a

simplified rectangular stress block, calculate the amount of reinforcement required given that fcu

= 30N/mm2 and fy = 410N/mm2.

b) Determine the amount of reinforcement using the design chart. Compare the two results.

Q7 Two beam sections are shown below, one being rectangular and the other flanged section.Both are used to carry a moment of 615kNm. Compare the areas of steel required for each case and hence give the advantages of flanged beams over rectangular ones.

24

All dimensions are in mm.

For both cases, fcu = 30N/mm2, fy = 410N/mm2, d = 618mm d1= 60mm (embedment of compression reinforcement) Use 0.72fy for compression steel and 0.87fy for the tension steel.

Q8A reinforced concrete beam cantilivering on one end carries a uniformly distributed load of 30kN/m as shown in figure1. The breadth and effective depth of the beam are 300mm and 400mm, respectively. Given the characteristic strength as fy =460N/mm2 for reinforcement and fcu = 30N/mm2 for concrete, determine:

(a) the maximum bending moments (sagging and hogging).Draw Bending moment diagram.

(b) the reinforcement for the beam.(c) the scheme of reinforcements.

Fig. 1.

Q9Determine the maximum moment of resistance at ultimate limit state and the of steel required for a rectangular beam given that b = 300mm, d = 500mm, fcu = 25N/mm2, fy = 410N/mm2.Q10A simply supported concrete lintel of effective span 4.5m is 450mm deep by 300mm wide.It is made of grade 30 concrete and is required give a 1 hour fire resistance in mild exposure. The lintel carries an imposed load of 15kN/mand dead load including self weight of 20kN/m. If bearing distance of the lintelis 150mmeach end, determine:(a)tensile reinforcement(b)shea reinforcement.Use fy = 460N/mm2,fyv = 250N/mm2 and diameter of the tensile steel reinforcement as 20mm.

Q11

(b) Determine the ultimate moment of resistance of the tee section shown in figure 1 if characteristic

material strengths are fy = 460N/mm2, fcu = 30N/mm2. Asumme s =hf.

25

Figure 1

Q12(b) A simply supported concrete beam of cross- section 250 450mm spans 3.0m. The dead load including self-weight of the beam is 8.0KN/m and the live load is 6.0KN/m. The material of construction are fcu = 30N/mm2, and fy = 460N/mm2.

Determine: -

(i) Ultimate design load(ii) The maximum flexural bending moment at mid span(iii) The area of steel at mid span if the effective depth of the beam is 400mm. Use design chart.

Q13A simply supported beam of the cross-section 300 500mm is doubly reinforced as shown in fig.1.Find the area of steel required in tension (As) and the area of steel in compression As’ in order to resist a dead load of 20KN/m and an imposed load of 20KN/m excluding self weight. Take the effective span as 6.92m, fcu = 40N/mm2 & fy = 460N/mm2.

Fig.1

Q14An RC beam of uniform cross-section is to carry an ultimate design load W = 10kN/m and a midspan point load of P = 40kN.The beam is freely supported with effective span of 6.0m.

(a) Determine the amount reinforcement required if b = 200mm, d = 350mm, fcu = 30N/mm2 and fy = 460N/mm2

(b) Check for deflection if the service stress is given as 288N/mm2

Q15A simply supported rectangular beam carries characteristic dead load gk of 10kN/m including the beam self weight. Characteristic imposed load carried by beam qk is 12kN/m.the beam has a breadth of 230mm and the effective depth d of 450mm. If the material strength are fcu = 30N/mm2 and fy = 460N/mm2,(a) Will the section be singly or doubly reinforced?(b) Determine the amount of reinforcing steel bars required.(c) Check the serviceability limit state of deflection for the beam.

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Q16A student working at a construction site on an industrial training investigated a beam whose section is shown in figure Q3 below. The beam has a span of 6.0m and can be considered to be simply supported. If the material strengths are fcu = 25N/mm2 and fy = 460N/mm2,

(a) determine the design load (b) calculate the characteristic imposed load when the characteristic dead load is 29.1kN/m(c) what condition must be satisfied for the beam to be doubly reinforced?

Q17A flanged beam section shown in figure Q4 is to support an ultimate moment M = 300kNm. The material strengths are fcu = 30N/mm2 and fy = 460N/mm2.

(a) Calculate the ultimate moment of resistance Mu of the section.(b) Check whether a simplified rectangular stress block lies below the flange or not.(c) Determine the amount of reinforcement required

Q18

A reinforced conrete beam is required to transmit an ultimate bending of 140kNm, inlclusive of its

own weight. Using the simplified stress block formulae given in BS8110, determine the depth of

beam required and the amount of steel needed in a 250mm wide beam for the following

combination grade 30 conrete with mild steel einforcement.

Q19

Show the variation of amount of reinforcement for a section of width 250mm and effective depth of

700mm if the design ultimate moment varies from 300kNm to 900kNm if it is known that

and .

27

d =

350m

m

300mm

hf =100mm

Fig 4

beff = 900mm

Q20

The singly reinforced concrete beam shown in figure 1 is required to resiet an ultimate moment of

550kNm. If the beam is composed of grade 30 concrete and high yield (HY) reinforcement. Check the

section size and determine the area of steel required.

Q21

(a) If it known that fcu = 25N/mm2,fy = 250N/mm2 and concrete cover is 50mm for the

cross section above, determine

(i) the depth, of the compression zone

(ii) the ultimate moment of resistance.

(b) If the bending moment to be applied to a rectangular cross section at the ultimate limit state is

300kNm, b = 300mm, h = 600mm, cover = 40mm,fcu = 25N/mm2 and fy = 460N/mm2,determine

the number and size of reinforcing bars.

Q22A rectangular beam carries a moment M = 550kNm. The beam dimensions are b = 300mm and d =

700mm. Using a ismplified rectangular stress block, determine the steel area required and the amount

of steel to be provided. Materials used : concrete fcu = 25 N/mm2,

Steel fy = 410N/mm2.

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Fig 1

As

b = 350mm

hd =600

Beam cross section

200mm

400mm

3R20

Q23Determine the amount of steel for the section to resist the given moment.

M = 175kNm, d = 400mm, d’= 40mm, fy = 250N/mm2, fcu = 25N/mm2, b = 230mm

Q24Design the section given in fig 24Given M = 500kNm fy = 425N/mm2

fcu = 25N/mm2

Q25

Determine the ultimate moment of resistance of the section given below using the chart for singly reinforced beams. fcu = 25N/mm2

fy = 460N/mm2

Concrete cover = 40mm

Q26A rectangular beam is to be reinforced with 4T20 and T10-300, but in the market there is only R20 and R8. Make necessary changes (if any) that will enable you to use the available materials (reinforcement). (Practical Question)

Q27A simply supported singly reinforced beam of 450mm deep by 250mm wide made of grade 30 concrete and required to give 2hrs fire resistance in mild exposure. The beam carries a characteristic dead load of 10kN/m (excluding its self-weight) and characteristic imposed load of 15kN/m.The beam frames in to 500mm square columns at each end.If T20 (fy = 460N/mm2) and R10 (fy = 250N/mm2) bars are used determine:-(a) Maximum effective span to the nearest metres for beam to carry the specified loads

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d

d’

b

200mm

400mm

Fig 24 600mm

1500mm

300mm

(b) Maximum clear span to the nearest millimetres(c) Maximum overall span to the nearest millimetres(d) The area of reinforcement provided to support the moment for beam whose effective span has been

determined (a){don’t use Grouped bar area table}

Q28The fig A below shows a beam made of grade 30 concrete required giving 2hrs fire resistance in mild exposure. The beam is designed to support brick wall with the density of 1500kg/m3 and characteristic imposed load of 10 kN/m. By using T20 and R10 design the beam fully.

Q29 A simply supported singly reinforced beam of 450mm deep by 250mm wide made of grade 30

concrete and required to give 3hrs fire resistance in mild exposure. The beam carries a characteristic dead load of 25kN/m (excluding its self-weight) and characteristic imposed load.The beam has an overall span of 4530mm and frames in to 450mm square columns at each end.If 4T20 (fy = 460N/mm2) and R10 (fy = 250N/mm2) bars are used, without using Grouped Bar Area Table determine; the maximum characteristic imposed load to be carried by the beam.

Q30 A simply supported T beam with an effective length of 3000mm and overall depth of

550mm is made of grade 30 concrete and is required to give 1.5hrs fire resistance in mild exposure. The beam carries a characteristic dead load of 50kN/m (excluding its self wt) and characteristic imposed load of 51.47kN/m. If the web width and flange thickness/depth of the beam are 250mm and 150mm respectively and the only reinforcements available in the market are T20 and R10 :-(a) Determine the area of reinforcements to be provided in the beam(checking for max and

min allowable %As may be disregarded) (b) Check for the deflection

30

230

300

450

varies

Fig A all dimensions are in mm

450 3100

A

A

2500

450

A-A450

The values of b are given in table 3.28 of BS 8110.Table 3.28 Values of bond coefficient bBar type Bar in tension Bar in compressionPlain bars 0.28 0.35Type1:deformed bar (Y) 0.40 0.50Type2:deformed bar (T) 0.50 0.63Fabric 0.65 0.81

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Beams design and analysis

Engineering

Transcript of Beams design and analysis