Unit 6 ( DESIGN OF REINFORCED CONCRETE CONTINUOUS BEAMS )

REINFORCED CONCRETE STRUCTURAL DESIGN

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UNIT 6 DESIGN

OF

REINFORCED

CONCRETE

CONTINUOUS BEAMS

OBJECTIVE

GENERAL OBJECTIVE To understand the principles of designing reinforced concrete continuous beams.

SPECIFIC OBJECTIVES

At the end of this unit you should be able to: 1.

define continuous beams.

2.

calculate total ultimate load of continuous beams.

3.

use moment and shear force coefficients given in BS 8110 to solve problems.

4.

calculate the area of reinforcement for continuous beams.

5.

provide the calculations for the total number and bar size for continuous beams.

6.

sketch reinforcement details for continuous beams.


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INPUT 1

6.1

Introduction

Continuous beams are used in structural designs when three or more spans exist. Continuous beams occur frequently in cast in situ construction when a single span of construction is linked to an adjoining span. Bending moment of continuous beams does not confine to a single span only but it will affect the whole system. It is most important that you develop a mental picture of the deflected form of the structure. If this can be visualized the areas of tension and compression reinforcement can readily be determined. The failure mode of continuous beams is shown in Figure 6.1 on the next page.

2T32 1T25

1T32 2T25

Figure 6.1: The reinforcement details

2T32 3T25

2T32 3T25

2T32 1T25

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Note that the sagging moment occurs at the middle of span while the hogging moment occurs at supports causing tension zone at the bottom and top of the beam section respectively. When sagging moment occurs, the beam and slab act together and hence the beam can be designed as having T- section. At supports, the beam must be designed as rectangular sections because hogging moment at this point can cause tension in the slab. Therefore, the tension reinforcement must be provided near the middle span i.e the bottom section (termed as bottom steel) and at top section (termed as top steel) at supports. Note that, to ensure continuity, reinforcement must be extended beyond the points that they are required.

6.2

Load Arrangement

For continuous beams, the load acting at a particular span is not necessarily the same as that acting at other spans. Therefore, the bending moments and shear forces at the supports and near the middle of the span will always vary according to the load they actually carry. This aspect should be taken into account when designing, by assigning the most severe loading case at any given section to give the maximum bending moment and shear force possible. The selection of the most critical load arrangements should be carried out according to clause 3.2.1.2.2, BS 8110.


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This is restated below; a)

all spans is to be loaded with the maximum design ultimate load i.e. 1.4 Gk + 1.6 Qk

b)

alternate spans are loaded with the maximum design ultimate load and all other spans loaded with the minimum design ultimate load , i.e. 1.0 Gk

There are three different methods that can be used to determine the bending moments and shear force. They are as follows: 

using bending moment and shear force coefficients given by BS 8110.



elastic analysis using moment redistribution



computer analysis

In this unit, only the first method (a) will be discussed. However you are encouraged to find for the other two methods if you are interested.

Now do the following activity.


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ACTIVITY 6a

Fill in the blanks with the correct answers. 6.1 When a single span of construction is linked to an adjoining span, this beam is designed as __________________________. 6.2 ___________________ span confine bending to that span alone, while bending of continuous beams affect the whole system. 6.3 The deflected forms of continuous beam resemble a __________________ moment near middle span and __________________ moment at supports. 6.4 The most ________________________ condition of loading is to be used in the design of continuous beams. 6.5 All spans are to carry _________________________ while alternate spans carry _________________________. 6.6 In this unit ________________ method is used to calculate bending moment and shear forces. 6.7 Tension

reinforcement

in

continuous

beams

is

denoted

as

___________________ steel and provided near middle span and ____________________ are to be provided at supports.


FEEDBACK 6a

ANSWERS

6.1 continuous beam 6.2 simply supported 6.3 sagging, hogging 6.4 critical 6.5 1.4 Gk + 1.6 Qk , 1.0 Gk 6.6 BS 8110 bending moment and shear force coefficients 6.7 bottom, top

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INPUT 2

6.8

Simplified analysis of continuous beams BS8110: Part1, Clause 3.4.3 allows the use of bending moment and shear force coefficients given in Table 3.6. They can only be used if the following conditions are as follows: •

the characteristic imposed load Qk does not exceed the characteristic dead load Gk.

• 3

the load is fairly uniformly distributed over three or more spans

the variation in the spans does not exceed 50% of the largest. Table 3.6, BS8110 is reproduced as below: at outer near middle at first at middle of at interior support of end span interior interior span supports support Bending Moment Shear

0

0.09F 

-0.11F 

0.07F 

-0.08F 

0.45F

-

0.60F

-

0.55F

Table 3.6: Design Ultimate Bending Moments and Shear Forces Design

Here F represents the total ultimate load on the span and number of redistribution of moments that is allowed.  is the effective span of the beam. It should be noted that in the case of a continuous beam the maximum moment occurs at the support and the section considered would therefore be a simple rectangular one as explained before. Now do the following activity.


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ACTIVITY 6b

6.7

A continuous beam with the following specification is shown below: -

Given: gk = 45 kN/m qk = 35 kN/m Calculate the following at each location along the span a) the total design ultimate load, f in kN. b) bending moments and c) shear forces 6.8

Complete the following table: -

at outer near the at the first at the middle at interior support middle of interior of interior supports end span support span Moment Shear


FEEDBACK 6b

Compare your answers that are given here. 6.7 a) Maximum distributed load. = 1.4gk + 1.6qk = 1.4(45) + 1.6(35) = 119 kN/m

b) F = 119 x span = 119 x 8.0 kN = 952 kN After you have this value, now calculate the moments as follows: Near the middle of end span (sagging) M = 0.09F  = 0.09 x 952 x 8 kNm = 685.44 kNm At the outer support, the moment is equal to zero.

At the first interior support (hogging)

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M = -0.11F  = -0.11 x 952 x 8 kNm = -837.76 kNm

At the middle of interior span (sagging): M = 0.07F  = 0.07 x 952 x 8 kNm = 533.12 kNm

At the interior supports (hogging): M = -0.08F  = -0.08 x 952 x 8 kNm = 609.28 kNm

c) Calculate the shear forces, V (kN) At outer support: V = 0.45F = 0.45 x 952 = 428.4 kN

Near the middle of end span: V=0 At the first interior support:

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V = 0.6F = 0.6 x 952 = 571.2 kN

At the middle of interior span: V=0

At the interior supports: V = 0.55F = 0.55 x 952 = 523.6 kN

6.8 Now complete the given table: -

Moment Shear

at outer near middle at first support of end span interior support 0 0.09F  -0.11F  0.45F 0.60F

INPUT 3

at middle of at interior interior supports spans 0.07F  -0.08F  0.55F


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6.3.1 Design example The arrangement of a continuous beam, is shown below (on plan): -

40 mm A

B 40 mm

5.0 m

5.0 m

5.0 m

The width of the beam is 300 mm and the overall depth is 660 mm. It has three equal spans of 5.0 m each. The beam is arranged at 4.0 m from centre to centre, while the thickness of the slab is 180 mm. The imposed live load q k on the beam is 50 kN/m and the dead load gk inclusive of self-weight of the beam is 85 kN/m. The characteristic strength of materials are fcu = 30 N/mm2, fy = 460 N/mm2 and fyv = 250 N/mm2. Assume that the mild exposure for the beam and the nominal cover of 25 mm.

Solution: For each span:


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The ultimate load, Wu = (1.4gk + 1.6 qk) kN/m = (1.4x 8.5 + 1.6 x 50 ) kN/m = 199 kN/m

Total ultimate load, F = 199 x 5.0 = 995 kN

Since; a. qk >gk b. equal span c. uniformly distributed load Therefore, the bending moment coefficients in Table 3.6, BS 8110 can be used. Bending a) Near the middle of the end span, the section is designed as T-beam. Moment, M = 0.09F  = 0.09 x 995 x 5 kNm = 448 kNm Effective flange width = bw +

0.7 L 5

= 300 +

0.7 x5000 mm 5

= 1000 mm K =

M bd 2 f cu


=

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448 ×10 6 1000 × (600 ) 2 × 30

= 0.041

Note that b is the effective flange width: Z = 0.95 d = 0.95 x 600 = 570 mm

d – Z = 600 – 570

=

hf 3 〈0 ) 2

This shows that the stress block is in the flange. As =

=

M 0.87 f y Z

448 ×10 6 0.87 × 460 × 570

= 1964 mm2 Use 2T32 bars together with one T25 bar (Total As = 2101 mm2) These bars are to be provided as bottom steel.


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b) At the interior supports, the beam is designed as a rectangular section. M = 0.11F  = 0.11 x 995 x 5 = 547 kNm (hogging moment)

M 547 ×10 6 = bd 2 f cu 300 × (580 ) 2 × 30

= 0.18〉 0.156 This shows that compression reinforcement is required. Note that in this calculation, b is 300 mm. i.e. the width of the web.

As ' =

=

M − 0.156 f cu bd 2 0.87 f y ( d − d ' ) 547 ×10 6 − 0.156 ×30 × 300 × (580 ) 2 0.87 × 460 × (580 − 50 )

= 352 mm2

This reinforcement is provided by extending the mid –span steel beyond the supports. As =

=

0.156 f cu bd 2 + As ' 0.87 f y Z 0.156 × 30 × 300 × (580 ) 2 + 352 0.87 × 460 × 0.775 × 580

= 2977 mm2 Note that Z = 0.775 d


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Use 2T32 bars together with 3T25 (Total As = 3080 mm2 and they are to be provided as top steel)

c) At the middle of interior span, a T-beam is designed M = 0.07F  = 0.07 x 995 x 5 = 348 kNm

As =

M 0.87 f y Z

348 ×10 6 = 0.87 × 460 × 0.95 × 600

= 1525 mm2 Use 1T32 together with 2T25, As = 1786 mm2 (bottom steel)

The reinforcement details at end span are shown in the sections below:-

T32

T25

T32

a) Mid-span details


T32

T25

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T32

T25

T25

T32

T25

T32

b) Near interior support details

SUMMARY


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1. Continuous beams occur in cast in-situ construction. 2. Arrangement of loads on continuous beams is to give maximum moments. 3. To obtain the shear forces and moments for design, Table 3.6, BS 8110 can be used. These values are subjected to the following conditions: 

the loading is uniformly distributed



the characteristic dead load is greater than the characteristic imposed load



the beam must consist of three or more than 15% of the longest span.

4. When sagging moment occurs near mid-span, the beam is designed as a Tbeam. 5. At supports where hogging moment occurs, the beam is designed as rectangular section 6. At the supports, top steel is provided. 7. Near the middle-span, steel is provided.

SELF-ASSESSMENT


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Using the given figure and the data provided, answer all the questions by encircling the answer of your choice. You may do all the calculations on a separate sheet of paper.

A

C

B

D

1 3m 2 3m 3 3m 4 A 3m

5 A 8m

8m

Data: Slab thickness = 125 mm Finishes, partitions etc. = 2.95 kN/m2 Characteristic imposed load = 3.00 kN/m2

8m


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Moderate exposure 2 hours fire resistance Characteristic concrete strength, fcu = 30 N/mm2 Characteristic steel strength, fy = 460 N/mm2 Beam dimensions bw x h = 250 x 450 mm All questions are associated with the design of continuous beam 3/A-D 1.

The effective flange width of beam 3/A-D is equal to …. A. 1170 mm B. 1270 mm C. 1370 mm D. 1570 mm

2.

Nominal cover to be used in the design is ….. A. 20 mm B. 25 mm C. 30 mm D. 35 mm

3.

Self-weight of the beam in kN/m is … A. 1.95 kN/m B. 2.95 kN/m C. 3.95 kN/m


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D. 4.95 kN/m 4.

The design load on beam 3/A-D is … A. 32.12 kN/m B. 42.12 kN/m C. 52.12 kN/m D. 62.12 kN/m

5.

Bending moments and shear forces for design is calculated using values in Table 3.6, BS 8110 because of the following conditions; A. the loading is uniformly distributed B. gk is greater than qk C. three equal spans D. all of the above

6.

The total ultimate load, F is equal to … A. 36.96 kN B. 136.96 kN C. 236.96 kN D. 336.96 kN

7.

If φ

bar

= 25 mm, φ

link

= 10 mm, the effective depth of the beam is .………. A. 365 mm B. 375 mm C. 385 mm


D. 395 mm

8.

The moments near the middle span A-B and C-D is …. A. 142.6 kNm B. 242.6 kNm C. 342.6 kNm D. 442.6 kNm

9.

Bending moment at support B and C is… A. 296.5 kNm B. 396.5 kNm C. 496.5 kNm D. 596.5 kNm

10.

Bending moment near the mid-span B-C is… A. 88.7 kNm B. 188.7 kNm C. 288.7 kNm D. 388.7 kNm

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11.

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Compression reinforcement is required at the following location; A. Near mid-span A-B and C-D B. At supports B and C C. Near mid-span B-C D. None of the above

12.

The area of reinforcement, As to be provided near mid-span A-B and C-D is … A. 1453 mm2 B. 1553 mm2 C. 1653 mm2 D. 1753 mm2

13.

The area of tension reinforcement required at supports B and C is … A. 2404 mm2 B. 2504 mm2 C. 2604 mm2 D. 2704 mm2

14.

The area of reinforcement, As required near mid-span B-C is …


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A. 890 mm2 B. 1089 mm2 C. 1189 mm2 D. 1289 mm2 15.

The following bars are to be used near mid-span B-C; A. 1T25 B. 2T25 C. 3T25 D. 4T25

FEEDBACK ON SELF-ASSESSMENT

Check your answers now! 1.

C

2.

C

3.

A


4.

B

5.

D

6.

D

7.

C

8.

B

9.

A

10.

B

11.

B

12.

C

13.

A

14.

D

15.

C

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What is your score in this test? To pass this unit, you should get 80% or more. If you have scored less than 80%, it is recommended that you study this unit again.

END OF THIS UNIT

Unit 6 ( DESIGN OF REINFORCED CONCRETE CONTINUOUS BEAMS )

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