aait, school of civil and environmental engineering reinforced concrete i · 2015-11-18 · aait,...

AAiT, School of Civil and Environmental Engineering Reinforced Concrete I

Examples on analysis of singly reinforced sections

Cross-Sectional analysis of reinforced concrete beam section for flexure

Prepared by: Concrete materials and structures chair

of 21

Example 2.1: Determine the design flexural strength of the beam cross – section shown

below. Use concrete class C20/25 and steel grade S-400

Step 1: Design Values ( Changing the characteristic value to design value)

400 25 8 5 362d mm

EN 1992

Formula 3.15

for persistent and transient design situation:

EN 1992

Table 2.1N

EN 1992

Section 3.1.6

NB: Take British National

Annex

Step 2: Assume the type of failure

Assume tension failure with rupture of steel and





of 21

Step 3:

Draw the strain profile corresponding to the type of failure and use the

similarity of triangles to develop a relationship between the unknown strain

and the neutral axis.

From Similarity of Triangle

25

25cm cm cm

x

cm

xk

d x d

(1)

EN 1992

Figure 6.1

NB: The

limiting value

for

is taken from

British National

Annex

Step 4: Use the equation of alpha corresponding to the assumption in step 2 and the

relationship developed in step 3 to calculate the unknown strain.

From equilibrium of forces,

But: and s s ydT A f

22 5 347.830.0444

11.33 300 362c

(2)

For 2cm c , 6

12cm

c cm xk

Substituting kx from Eqn (1),





of 21

6

12 25cm cm

c cm

cm

2 36

12 300cm cm

c

cm

2 312 300 6c cm c cm cm

3 26 12 300 0cm cm c cm c

3 26 12* 0.0444 * 300 * 0.0444 0cm cm cm

3 26 0.5328 13.32 0cm cm cm

Solving the cubic equation results three possible answers

1 5.4546cm > 3.5 …………not ok!

2 1.3136cm < 0 …………..not ok!

3 1.859cm < 3.5 ………….ok!

Thus, 1.859cm ‰

Step 5: Check if the assumption in step 2 is correct and if it is, proceed to step 8. If

the assumption is not correct, repeat step 2 to 5 with another assumption.

1.859 2 !cm the assumptionsarecbotho orr ctf e

Step 6: Calculate the value of beta

For 2cm c , 8

4 6cm

c x

cm

k

1.8590.0692

25 1.859 25cm

x

cm

k and

25.0504x mm

Substituting the values of kx and εcm yields , βc = 0.0257

Step 7: Calculate the moment resistance

1s yd cM A f d

22* 5 347.83 362 1 0.0257 19.27M kNm





of 21

Example 2.2: Repeat Example 2.1 replacing 2 Ø 10 by 2 Ø 12


400 25 8 6 361d mm

EN 1992

Formula 3.15


EN 1992

Table 2.1N

EN 1992

Section 3.1.6

NB: Take

British National

Annex







of 21

Step 3: Draw the strain profile corresponding to the type of failure and use the




25

25cm cm cm

x

cm

xk

d x d

(1)

EN 1992

Figure 6.1

NB: The

limiting value

for

is taken from

British National

Annex

Step 4: Use the equation of alpha corresponding to the assumption is step 2 and the




22 6 347.830.064

11.33 300 361c

(2)

For 2cm c , 6

12cm

c cm xk


6

12 25cm cm

c cm

cm





of 21

2 36

12 300cm cm

c

cm

2 312 300 6c cm c cm cm

3 26 12 300 0cm cm c cm c

3 26 0.768 19.2 0cm cm cm

Solving the cubic equation results three possible answers

1 5.116cm > 3.5 …………not ok!

2 1.545cm < 0 …………..not ok!

3 2.429cm < 3.5 ………….ok!

Thus, 2.429cm ‰



2.429 3. !5cm the assumption iscorrect

2.429 2 !cm the assumption isnot correct

Trial 2


For 2cm c , 3 2

3cm

c x

cm

k

Substituting kxfrom Eqn (1),

3 2

25 3cm cm

c

cm cm

3 75 3 2c cm c cm

2 75 2 75 * 0.0642.42

3 3 3 * 0.064 3c

cm

2.429 2 !cm the assumption iscorrect

2.429 3. !5cm the assumption iscorrect





of 21

Step 6:

Calculate the value of beta

For 2cm c ,

3 4 2

2 3 2

cm cm

c x

cm cm

k

2.420.088

25 2.42 25cm

x

cm

k and

31.826x mm



1s yd cM A f d

22* 6 347.83 361 1 0.0343 27.43M kNm





of 21



400 25 8 7 360d mm

EN 1992

Formula 3.15


EN 1992

Table 2.1N

EN 1992

Section 3.1.6

NB: Take

British National

Annex







of 21

Step 3:





25

25cm cm cm

x

cm

xk

d x d

(1)

EN 1992

Figure 6.1

NB: The

limiting value

for

is taken from

British National

Annex

Step 4: Use the equation of alpha corresponding to the assumption is step 2 and the




24 7 347.830.175

11.33 300 360c

(2)

For 2cm c , 3 2

3cm

c x

cm

k


3 2

25 3cm cm

c

cm cm





of 21

3 2

3 75cm

c

cm

3 75 3 2c cm c cm

2 75 2 75 * 0.1756.11

3 3 3 * 0.175 3c

cm



6.11 3. !5cm the assumption isnot correct

Trial 2

Assume tension failure with crushing of concrete


3.5 3.5 3.5

3.5s

x

s

xk

d x d

(1)

For 2cm c , 3 2

3cm

c x

cm

k


3 23.5

3.5 3cm

c

s cm

3.5 3 * 3.5 2

3.5 3 * 3.5c

s





of 21

8.5

10.5 3c

s

10.5 3 8.5c s c

8.5 10.5 8.5 10.5 * 0.17512.69

3 3 * 0.175c

s

c

3

347.831.74

200 10

yd

yd

s

f

E

12.69 1.74 !s the assumption iscorrect

12.69 25 !s the assumption iscorrect


For 2cm c ,

3 4 2

2 3 2

cm cm

c x

cm cm

k

3.5 3.50.216

3.5 3.5 12.69x

s

k and

77.76x mm



1s yd cM A f d

24 * 7 347.83 360 1 0.09014 70.15M kNm





of 21



400 25 8 12 355d mm

EN 1992

Formula 3.15


EN 1992

Table 2.1N

EN 1992

Section 3.1.6

NB: Take

British National

Annex


Assume tension failure with crushing of concrete





of 21

Step 3:





3.5 3.5 3.5

3.5s

x

s

xk

d x d

(1)

EN 1992

Figure 6.1

Step 4: Use the equation of alpha corresponding to the assumption in step 2 and the

relationship developed in step 3 to calculate the unknown strain



25 12 347.830.652

11.33 300 355c

(2)

For 2cm c , 3 2

3cm

c x

cm

k

Substituting kxfrom Eqn (1),

3 23.5

3.5 3cm

c

s cm





of 21

3.5 3 * 3.5 2

3.5 3 * 3.5c

s

8.5

10.5 3c

s

10.5 3 8.5c s c

8.5 10.5 8.5 10.5 * 0.6520.846

3 3 * 0.652c

s

c



3

347.831.74

200 10

yd

yd

s

f

E

0.846 !s yd the assumption isnot correct

Trial 2

Assume compression failure with crushing of concrete


3.5 3.5 3.5

3.5s

x

s

xk

d x d

(1)

For 2cm c , 3 2

3cm

c x

cm

k





of 21

2 3 35 12 10 200 *100.375

11.33 300 355

s

c s

(2)

For 2cm c , 3 2

3cm

c x

cm

k


3 23.5

3.5 3cm

c

s cm

3.5 3 * 3.5 2

3.5 3 * 3.5c

s

8.50.375

10.5 3c s

s

210.5 * 0.375 3 * 0.375 8.5s s

21.125 3.9375 8.5 0s s

Solving the quadratic equation results two possible answers

1 1.508s > 0 ………...…ok!

2 5.008s < 0 …………..not ok!

Thus,

1.508 1.74 !s the assumption iscorrect





of 21

Step 6:

Calculate the value of beta

For 2cm c ,

3 4 2

2 3 2

cm cm

c x

cm cm

k

3.5 3.50.699

3.5 3.5 1.508x

s

k and

248.1x mm



1s yd cM A f d

25 *12 347.83 355 1 0.2912 171.31M kNm





of 21

Example 2.5: Find the reinforcement amount which results balanced failure &

calculate the moment capacity for the cross section described in Example 2.1. (Assume

the effective depth to be 354 mm)


EN 1992

Formula 3.15


EN 1992

Table 2.1N

EN 1992

Section 3.1.6

NB: Take

British National

Annex

Step 2: Strain profile for balanced failure

From similarity of triangles

3.5 yd

b bx d x

but 1.74yd





of 21

3.5 1.74

b bx d x

3.50.668

3.5 1.74xk

236.45x mm

For 2cm c , 3 2

3cm

c x

cm

k

3 * 3.5 20.668 0.5408

3 * 3.5c

Step 3: Evaluating sA

s s ydT A f

0.508 *11.33 * 300 * 354

347.83sA

21870.79sA mm


For 2cm c ,

3 4 2

2 3 2

cm cm

c x

cm cm

k



1s yd cM A f d

1870.79 347.83 354 1 0.278 166.35M kNm





of 21

Example 2.6: for the cross section described in Example 2.1, Find the reinforcement

amount & calculate the moment capacity if kx = 0.448. (Assume the effective depth to

be 354 mm)


EN 1992

Formula 3.15


EN 1992

Table 2.1N

EN 1992

Section 3.1.6

NB: Take

British National

Annex

Step 2: Strain profile

From similarity of triangles

3.5 s

x d x

but 1.74yd





of 21

3.5 1 13.5 1 3.5 1 4.3125

0.448s

x

d x

x k

For 2cm c , 3 2

3cm

c x

cm

k

3 * 3.5 20.448 0.363

3 * 3.5c

Step 3: Evaluating sA

s s ydT A f

0.363 *11.33 * 300 * 354

347.83sA

21255.72sA mm


For 2cm c ,

3 4 2

2 3 2

cm cm

c x

cm cm

k



1s yd cM A f d

1255.72 347.83 354 1 0.1864 125.8M kNm



Summary

x (mm) cm ‰ s ‰ Mmax

Example 2.1 As = 157 mm2 25.05 1.86 25 19.27

Example 2.2 As = 226.08 mm2 31.83 2.42 25 27.43

Example 2.3 As = 615.44 mm2 77.76 3.5 12.65 70.15

Example 2.4 As = 2260.8 mm2 248.1 3.5 1.51 171.31

Example 2.5 As = 1870.79 mm2 236.45 3.5 1.74 166.35

Example 2.6 As = 1255.72 mm2 158.59 3.5 4.31 125.8

aait, school of civil and environmental engineering reinforced concrete i · 2015-11-18 · aait,...

Documents