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Lectures of Structure II 2017-2018[Department of architectural engineering]

Flexure in Beams

1. Introduction

There are several terms which need to be defined in this stage:

1.1. Type of Loads

To design any structural member, there are several kinds of load which should be taken into account by designers. Among these kinds the following:

1. Dead load

Dead loads are permanent or stationary loads which are transferred to structure throughout the life span. Dead load is primarily due to self-weight of structural members, permanent partition walls, fixed permanent equipment and weight of different materials. It majorly consists of the weight of roofs, beams, walls and column etc. which are otherwise the permanent parts of the building.

2. Live load

Live loads are either movable or moving loads without any acceleration or impact. These loads are assumed to be produced by the intended use or occupancy of the building including weights of movable partitions or furniture etc..

1.2. Methods of design:

1. Working Stress Design :WSD

2. Strength Design Method : SDM

2. Strength Design Method: SDM

Advantage of SDM over WSD:

1) Consider mode of failure

2) Nonlinear behavior of concrete

3) More realistic F.S.

4) Ultimate load prediction ≅ 5%

5) Saving (lower F.S.)

2.1. Load Factors

Factored Load= Load Factors × Service load

Dead Load Factor = 1.2

Live Load Factor = 1.6

1University of Babylon


-Other type of loads

Wind Load: load factor= 1.6

Lateral Earth Pressure: load factor =1.6

-Load Combinations

Factored Load = 1.2 DL + 1.6 LL

2.2. Strength Reduction Factors

Nominal Strength (N) = Strength of a member calculated using Strength Design Method

Strength Reduction Factor = factor that accounts for

(1) Variations in material strengths and dimensions

(2) Inaccuracies in the design equations

(3) Degree of ductility and required reliability of member

(4) Importance of member in the structure

Bending φ = 0.90 or determine by Equ. As in the next lecture

Shear and Torsion φ = 0.75

Compression φ = 0.65 (column with ties) or 0.7 (column with spirals)

φMn= Mu , φVn=Vu , φPn=Pu where:

n = Nominal strength

u = Ultimate load

2.3. Rectangular section with singly reinforcement

Singly reinforced section means that the section is subject to bending moment only.

Many necessary terms must be known before the start of design:

1-The effective depth (d) is the difference between the overall depth and the distance from the concrete cover to the center of gravity of the reinforced steel as in figure 1.

2-The main steel is the steel reinforcement which is necessary to resist the tension according to the bending moment as shown in figure 1.

3-The concrete cover as in figure 1 is necessary to protect the main steel which can be calculated according to codes or see table 1. The concrete cover depends on:

a) Environment conditions.


Figure 1.


b) Size of main reinforced steel bars.

c) Type of structure.

Table 1: Minimum concrete cover

Case Min. cover (mm)

Concrete cast against and permanently exposed to earth

75

Concrete exposed to earth or weather

No. 19 through No. 57 bars 50

No. 16 bar, MW 200 or MD 200 wire and smaller

40

Concrete not exposed to weather

or in contact with ground

Slabs, walls and joists No. 43 and No. 57 bars

40

No. 36 bar and smaller

20

Beams and columns 40

From the practical we can use concrete cover depend on the type of structure as follows:

a) Beam subject to moderate conditions, take cover 40 mm.

b) Foundation exposed to earth permanently, take cover 75 mm.

c) Slab exposed to earth permanently, take cover 20 mm.

2.3.1. Behavior of Concrete Beam

a-Before crack b- After crack



Equivalent Stress Distribution (Whitney stress block)

∑ Fx=0 C=T

0.85 f 'c∗ab=A s f y

a=A s f y

0.85 f 'c b

=ρ f y d

0.85 f 'c

Nominal Moment Strength ( Mn )

M n=T (d−a2 )=A s f y (d−

ρ f y d2 (0.85 ) f '

c)

M n=ρ f y b d2(1−ρ f y

1.7 f 'c)

For f 'c ≤ 30 MPa , β1=0.85

For f 'c>30 MPa, β1=0.85−0.05( f '

c−307 )≥ 0.65

cd=

εcu

εcu+ε y

c=( ε cu

εcu+ε y)d


f'cf'cf'cf'cf'cf'cf'cf'cf'cf'cf'cf'c


2.3.2. Limits of reinforcement ratio

1- Balanced condition (ρb)

Ideal condition as when the stress in the steel reaches fy and stress in concrete reaches max strength at the same time which called balance condition

[∑ F x=0¿¿C=T ;

0.85 f 'c β1cb=A s f y=ρb f y bd

ρb=0.85 f '

c

f yβ1( εcu

ε cu+ε y)

Substitute εcu=0.003 , ε y=f y

2.0 ×105

ρb=0.85 f '

c

f yβ1( 600

600+ f y )ρ=ρb :balance , ρ> ρb: overRC , ρ<ρb :u nderRC

2- Maximum reinforcement ratio (ρmax)

To make sure that the actual reinforcement ratio ρ ≤ ρb then the codes limit the reinforcement ratio by maximum ratio of reinforcement

ρmax=0.85 β1

f c 'f y

ϵ cu

ϵ cu+0.004

If the ρ >ρmax it leads to sudden failure and the section is called over reinforced section.

3- Minimum reinforcement ratio (ρmin)

To make sure that the cracks will not appear on the building you must take the reinforcement ratio not less than the minimum value that is:

Minimumsteelratio : ρmin=max(¿ 1.4f y

, √ f c '4 f y

) (concrete first crack )¿

4− ρt=0.85 β1

f c 'f y

ϵ cu

ϵ cu+0.005

2.3.3. Reduction Factor (φ¿

φ=0.9 if ρ ≤ ρt



φ=0.483+83.3 ϵ t>0.65 if ρ>ρt

where ρt=0.85 β1

f c 'f y

ϵ cu

ϵ cu+0.005, ϵ t=

ϵ cu∗d t−cc

,∧d t ≈ d

2.3.4. Design Procedure for Section with Tension Reinforcement onlyI. Unknown Dimensions and reinforcement

*If the self weight of the beam is unknown, you can start from step 1 etherwise you can go to step 3.

First step: (to determine self weight of beam) Select d from the recommended h

where d = h – cover – stirrup diameter -0.5 main reinforcement diameter

Second step: Find b where (d / b ≈1.5 to 2)

Third step: Find Wu

Wu = 1.2 D.L + 1.6 L.L

Fourth step: For statically determinate member draw bending moment diagram and find Mu. For continuous beam find Mu as in table s2.

Table 2: Moment Coefficients

Case Positive moment

At end span with discontinuous unrestrained end WuL2/11

At end span with discontinuous integral with support end WuL2/14

At interior span WuL2/16

Case Negative moment

At exterior face of first interior support for two spans WuL2/9

At exterior face of first interior support for more than two spans WuL2/10

At other faces of interior support WuL2/11

Fifth step: assume ρ between ρmax ≥ ρ ≥ ρmin ¿(preferred less than ρt to take ¿0.9 )



ρmin=max (¿ 1.4f y

, √ f c '4 f y

)¿

ρmax=0.85 β1

f c 'f y

ϵ cu

ϵ cu+0.004

β1={ 0.85 if ( f 'c ≤ 30 MPa)

0.85−0.05( f 'c−30

7 ) if (30< f 'c ≤56)

0.65 if (f 'c>56 MPa)

Sixth step: Compute dimensions of beam

from:M u=φρbd2 f y (1−0.59 ρf y

f 'c)

by find bd2 (assume d/b=2)

Eighth Step: Select steel reinforcement

A s=ρbd

Ninth step: check strength of section

M u≤ φρb d2 f y (1−0.59 ρf y

f 'c)

II. Known dimensions and unknown reinforcement

First step: For statically determinate member draw bending moment diagram and find Mu. For continuous beam find Mu as in table 2.

Second step: Find ρmax∧ρmin


, √ f c '4 f y

)¿

ρmax=0.85 β1

f c 'f y

ϵ cu

ϵ cu+0.004

β1={ 0.85 if ( f 'c ≤ 30 MPa)

0.85−0.05( f 'c−30

7 ) if (30< f 'c ≤56)

0.65 if (f 'c>56 MPa)

Third step : assume φ=0.9 ¿be ˇlater



Fourth step: Compute ρ

from:M u=φρbd2 f y (1−0.59 ρf y

f 'c)

and check ρmax ≥ ρ ≥ ρmin

Fifth step: Check φ by determine ρt

φ=0.9 if ρ ≤ ρt

φ=0.483+83.3 ϵ t>0.65 if ρ>ρt

where ρt=0.85 β1

f c 'f y

ϵ cu

ϵ cu+0.005, ϵ t=

ϵ cu∗d t−cc

,∧d t ≈ d

Sixth step: Check φMn≥ Mu

Example 1: A beam has a simply support span length of 8m and a rectangular cross section. Design a suitable section with singly reinforcement to resist the following load Wu = 10 kN/m (including the section self-weight)? Use f'c = 25 MPa and fy = 350 MPa.

Solution:

Mu = (10 x 82 / 8) = 80 kN.m

ρt=0.85 β1

f c 'f y

ϵ cu

ϵ cu+0.005=0.0193

Let ρ=0.016 thenφ=0.9

M u=φρbd2 f y (1−0.59 ρf y

f 'c)

80∗106=0.9∗0.016∗b d2∗350(1−0.59∗0.016∗35025

)

b d2=18.3∗106 mm3

Let d=340 mm

b=18.3∗106

3402 =160 mm


, √ f c '4 f y

)=max ( 1.4350

=0.004 , √254∗350

=0.0036)¿

¿0.004< ρ∴O. K .



ρmax=0.85 β1

f c 'f y

ϵ cu

ϵ cu+0.004

ρmax=0.85∗25

3500.85( 0.003

0.003+0.004 )=0.0221> ρ∴O. K .

A s=ρbd=0.016∗160∗340=870 mm2

Use 2∅ 25

A s(Provided )=980 mm2∴O . K .

Example 2: Design B1 in the floor plan shown below.

Slab thickness = 12 cm

LL = 3 kN/m2

f'c= 28 MPa

Steel: fy=400 MPa

Solution: Slab DL = 0.12(24) = 2.88 kN/m2

Ultimate load = 1.2(2.88) + 1.6(3) = 8.26kN/m2

Load on B2=[ 4∗22

× 8.26+( 4+12 )1.5 ×8.26 ]

4=16 kN /m

Assume width (b) of B2=0.3m and d=4/16 =0.4m

Then weight of B2=0.3*0.4*24*1.2 =3.46 kN/m

Reactions at B2's ends=wL/2=(16+3.46)*4/2=38.91 kN



Assume b=0.3 m and d=8/16 =0.5m

Weight of B1=1.2 *24* 0.3*0.5 =4.32 kN/m

Mmax=234.1 kN.m

use bar of 25mm diameter

for flexural reinforcement.

And let ds=10mm

d=500-40-10-25/2

=437.5 mm

ρb=0.85 f '

c

f yβ1( 600

600+ f y )ρb=

0.85∗28400

0.85( 600600+400 )

=0.0303

ρmax=0.75 ρb

¿0.75∗0.0303=0.0228

M u=φ ρ f y b d2(1− ρ f y

1.7 f 'c)

234.1∗106=0.9∗ρ∗400∗300∗437.52(1− ρ∗4001.7∗28 )


38.91 kN

18.17kN/m 14.46kN/m

38.91 kN

6.2 8.26kN/m

9.91 kN/m

B2=38.91 kN

kN.m

48.99-10.08- kN

234.1 -105.6

102.82

38.91kN

18.8722.58


ρ=0.0127

ρmin=1.4f y

= 1.4400

=0.0035

∴ ρmax (¿0.0229 )≤ ρ (¿0.0127 ) ≤ ρmin (¿0.0035 ) O. K .

A s=ρb d2=0.0127∗300∗437.5=1666.9 mm2

Use 4 φ 25 A s=1963 mm2

bmin=4∗25+3∗25+2∗10+40∗2=275 mmO . K .



HW (1)P1A rectangular simply supported concrete beam of width (b=300mm) is limited by architectural considerations to a total depth (h =600 mm). It must carry a total factored load (Wu =65.5 kN/m). Allow 50 mm to the center of the bars from the tension face of the beam. Material strengths are (fy= 400MPa and f'c=28 MPa). Use As=3500mm2 .Determine maximum allowable length of beam to resist applied load.P2A rectangular tension-reinforced beam is to be designed for dead load of 7.5kN/m (including self-weight) and service live load of 17.5kN /m, with a 6.7 simple span. Material strengths will be fy=420 MPa and f'c=21 MPa for steel and concrete, respectively. The total beam depth= 400 mm. Calculate the required beam width and tensile steel requirement, using a reinforcement ratio of 0.6ρmax. The effective depth may be assumed to be 67.5mm less than the total depth


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