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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
Page 1 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
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
Page 2 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),
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
Page 3 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
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
Page 4 of 21
Example 2.2: Repeat Example 2.1 replacing 2 Ø 10 by 2 Ø 12
Step 1: Design Values ( Changing the characteristic value to design value)
400 25 8 6 361d 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
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
Page 5 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 is 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 6 347.830.064
11.33 300 361c
(2)
For 2cm c , 6
12cm
c cm xk
Substituting kx from Eqn (1),
6
12 25cm cm
c cm
cm
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
Page 6 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 ‰
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.
2.429 3. !5cm the assumption iscorrect
2.429 2 !cm the assumption isnot correct
Trial 2
Assume tension failure with rupture of steel and
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
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
Page 7 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
Substituting the values of kx and εcm yields , βc = 0.0343
Step 7: Calculate the moment resistance
1s yd cM A f d
22* 6 347.83 361 1 0.0343 27.43M kNm
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
Page 8 of 21
Example 2.3: Repeat Example 2.1 replacing 2 Ø 10 by 4 Ø 14
Step 1: Design Values ( Changing the characteristic value to design value)
400 25 8 7 360d 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
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
Page 9 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 is 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
24 7 347.830.175
11.33 300 360c
(2)
For 2cm c , 3 2
3cm
c x
cm
k
Substituting kx from Eqn (1),
3 2
25 3cm cm
c
cm cm
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
Page 10 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
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.
6.11 3. !5cm the assumption isnot correct
Trial 2
Assume tension failure with crushing of concrete
From Similarity of Triangle
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
Substituting kx from Eqn (1),
3 23.5
3.5 3cm
c
s cm
3.5 3 * 3.5 2
3.5 3 * 3.5c
s
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
Page 11 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
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 3.50.216
3.5 3.5 12.69x
s
k and
77.76x mm
Substituting the values of kx and εcm yields , βc = 0.09014
Step 7: Calculate the moment resistance
1s yd cM A f d
24 * 7 347.83 360 1 0.09014 70.15M kNm
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
Page 12 of 21
Example 2.4: Repeat Example 2.1 replacing 2 Ø 10 by 5 Ø 24
Step 1: Design Values ( Changing the characteristic value to design value)
400 25 8 12 355d 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 crushing of concrete
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
Page 13 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
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
From equilibrium of forces,
But: and s s ydT A f
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
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
Page 14 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
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.
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
From Similarity of Triangle
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
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
Page 15 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
Substituting kx from Eqn (1),
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
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
Page 16 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
Substituting the values of kx and εcm yields , βc = 0.2912
Step 7: Calculate the moment resistance
1s yd cM A f d
25 *12 347.83 355 1 0.2912 171.31M kNm
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
Page 17 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)
Step 1: Design Values ( Changing the characteristic value to design value)
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: Strain profile for balanced failure
From similarity of triangles
3.5 yd
b bx d x
but 1.74yd
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
Page 18 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
Step 4: Calculate the value of beta
For 2cm c ,
3 4 2
2 3 2
cm cm
c x
cm cm
k
Substituting the values of kx and εcm yields , βc = 0.278
Step 5: Calculate the moment resistance
1s yd cM A f d
1870.79 347.83 354 1 0.278 166.35M kNm
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
Page 19 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)
Step 1: Design Values ( Changing the characteristic value to design value)
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: Strain profile
From similarity of triangles
3.5 s
x d x
but 1.74yd
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
Page 20 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
Step 4: Calculate the value of beta
For 2cm c ,
3 4 2
2 3 2
cm cm
c x
cm cm
k
Substituting the values of kx and εcm yields , βc = 0.1864
Step 5: Calculate the moment resistance
1s yd cM A f d
1255.72 347.83 354 1 0.1864 125.8M kNm
AAiT, School of Civil and Environmental Engineering Reinforced Concrete I
Examples on analysis of singly reinforced sections
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