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Flexural Analysis of Beams
1
Acknowledgement
This Powerpoint presentation was prepared by Dr. Terry Weigel, University of Louisville.
SI units are added by Dr. Tarek Ragab, University at Buffalo (SUNY)
2
Introduction
3
Gradual application of load on a simply supported beam until it fails
Three behavior stages:
(a) Concrete uncracked (b)Concrete cracked – elastic stresses (c) Concrete cracked – Ultimate strength
Uncracked Concrete Stage
4
Tensile stresses in concrete are less than modulus of rupture
Entire beam cross-section resists bending
Compression on one side and tension on the other
The effect of reinforcing steel is negligible on the beam properties
Uncracked Concrete Stage
5
Concrete Cracked – Elastic Stresses
6
Cracks develop on tension side of beam
Moment at which cracks begin – cracking moment (Mcr)
Cracks spread toward the neutral axis
Neutral axis moves upward (for positive moment)
Concrete Cracked – Elastic Stresses
7
Concrete Cracked – Elastic Stresses
8
Cracked concrete cannot resist tension
Tensile stress must be resisted by steel
This stage continues until concrete stresses are about one-half concrete compression strength and steel stress is less than yield
Stress varies linearly with strain
Concrete Cracked – Elastic Stresses
9
This is the stage occurring under service-load conditions
Stresses are computed using the transformed area method
Service or working loads are considerably higher than the cracking load
Concrete Cracked – Ultimate Strength Stage
10
Tensile cracks and neutral axis move upward (for positive moment)
Stresses are no longer linearly related to strain
Reinforcing bars yield
Concrete Cracked – Ultimate Strength Stage
11
Moment-Curvature Relationship
12
Cracking Moment
13
Effect of reinforcement is negligible until concrete is cracked
Stress in the beam may be calculated using:
g
Myf
I
ACI Section 9.5.2.3 – cracking moment is calculated using ACI equation 9-9:
r g
cr
t
f IM
y
Cracking Moment
14
Cracking stress – ACI Equation 9-10:
'0.62r cf f
yt is the distance from the centroidal axis to the extreme tension fiber
is a parameter to account for lightweight concrete:
= 1 for normal weight concrete = 0.85 for sand-lightweight concrete
= 0.75 for all-lightweight concrete
Example 2.1
15
For the beam shown in the figure, compute the bending stresses for a moment of 32 KN.m. Use normal weight concrete, a concrete strength of 28MPa and a modulus of rupture of 3.3MPa and determine the cracking moment.
Example 2.1
16
Example 2.1
17
(a) Calculate the bending stresses assuming the section uncracked
33 4
4
1 1300 450 2.278 9 mm
12 12
32 6 . (225 )3.15
2.278 9 mm
g
g
I bh mm mm E
E N mm mmMyf MPa
I E
Example 2.1
18
(b) Calculate the cracking moment
43.3 2.278 9 mm
225
33.41 6 . 33.41 .
r g
cr r
MPa Ef IM f S
y mm
E N mm KN m
Elastic Stresses - Cracked Concrete
19
All concrete in the tensile zone is cracked and is neglected
Perfect bond between the tension steel and concrete – strains in two materials are equal
Stresses are not equal because of differences in moduli
Elastic Stresses - Cracked Concrete
20
Ratio of the steel to concrete stress is given by the modular ratio
s
c
En
E
Area of tension steel (As) is equivalent to an equivalent concrete area of nAs
Transformed area
Elastic Stresses - Cracked Concrete
21
Figure 2.6
Example 2.2
22
Calculate the bending stresses in the shown beam using the transformed area method. Use a concrete strength of 21MPa, n = 9 and M = 90KN.m.
Example 2.2
23
Example 2.2
24
(a) Locate the neutral axis by summing moments of areas about the neutral axis
2
2
mm300 mm 9 1935 mm 425 mm
2
150 7.4 6 17,415
164
xmm x mm x
x E x
x mm
171 mm – Consider this value in next calculations
Example 2.2
25
(b) Compute the moment of inertia of the transformed area
3 22
4
1 164300 164 (300)(164)( ) (17,415) 425 164
12 2
1.627 9 mm
crI
E
where Icr is the cracked, transformed moment of inertia
Example 2.2
26
(c) Compute the bending stresses
4
4
(90 6 . ) 1648.84
1.67 9 mm
90 6 . 425 164( )9
1.67 9 mm
126.6
c
cr
s
cr
E N mm mmMxf MPa
I E
E N mmM d xf n
I E
MPa
Example 2.3
27
Determine the allowable bending moment that may be applied to the beam of Example 2.2 if the allowable stresses are 8MPa for concrete in compression and 140MPa for reinforcing steel in tension.
Example 2.3
28
4
4
8 1.627 9 mm79 6 .
164 mm
79KN.m
140 1.627 9 mm97 6 .
( ) 9 425 164
97 .
c crc
s crt
MPa Ef IM E N mm
x
MPa Ef IM E N mm
n d x
KN m
The beam capacity is controlled by the concrete stress and is 79KN.m
Example 2.4
29
Calculate the bending stresses in the notched beam shown using the transformed area method. Use n = 8 and M = 150KN.m.
Example 2.4
30
Example 2.4
31
(a) Locate the neutral axis (NA) by summing moments of areas about the neutral axis – assume the NA below the notch
2
2
mm450 mm 150 150 mm 75
2
8 3,276 mm 575mm mm
225 22,500 1.6875 6 15.0696 6 26,208
235
xmm x mm mm x mm
x
x x E E x
x mm
Note that the NA is below the depth of the slot, as assumed. If x had been less than 150 mm, the calculated value of x would not have been valid.
Example 2.4
32
(b) Compute the moment of inertia of the transformed area
3 3
22 4
1 1150 235 2 150 85
3 3
8 3,276mm 575 235 4.358 9 mm
crI
E
Example 2.4
33
(c) Compute the bending stresses
4
4
150 6 . 2358.1
4.358 9 mm
150 6 . 575 235( )8
4.358 9 mm
93.6
c
cr
s
cr
E N mm mmMxf MPa
I E
E N mmM d xf n
I E
MPa
Doubly-Reinforced Beams
34
Compression steel
Smaller beams
Reduction of long-term deflection
Stress in compression bars doubles over time
Support for stirrups
Example 2.5
35
Calculate the bending stresses in the beam using the transformed area method. Use n = 10 and M = 165KN.m.
Example 2.5
36 As time goes by, stress in compression steel is assumed to be doubled; 2nAs
Example 2.5
37
(a) Locate the neutral axis by summing moments of areas about the neutral axis
2
2
2
350 1,290 mm 20 1 502
10 2,580 mm 450
175 24,510 1.2255 6 11.61 6 25,800
139
xx x
x
x x E E x
x mm
10.38E6
Example 2.5
38
(b) Compute the moment of inertia of the transformed area
3 22
22 4
1350 139 20 -1 1,290 mm 89 mm
3
10 2,580 mm 311 mm 3 9 mm
crI
E
Example 2.5
39
(c) Compute the bending stresses
4
'
4
4
165 6 . 1397.65
3 9 mm
165 6 . 139-50( ')2 20 98
3 9 mm
165 6 . 450 139( )10 171
3 9 mm
c
cr
s
cr
s
cr
E N mm mmMxf MPa
I E
E N mmM x df n MPa
I E
E N mmM d xf n MPa
I E
Ultimate Moment
40
Tensile bars are stressed to yield before concrete crushes
b1 depends on concrete strength
Constant concrete compressive stress – 85% of ultimate stress
Rectangular stress block extends b1 times c
Whitney
stress
block
Example 2.6
41
Calculate the nominal flexural strength of the beam if the yield stress of the steel is 420MPa and the strength of the concrete is 21MPa.
Example 2.6
42
Example 2.6
43
(a) Calculate the tensile force (T) and the compressive force (C)
2
'
1935 420 812,700
0.85 0.85 21 350 6,247.5
s y
c
T A f mm MPa N
C f ab MPa a a
(b) Determine a by equating T and C
812,700 6,247.5
130
N a
a mm
Example 2.6
44
(c) Compute the moment arm and the moment
130525 460
2 2
( ) 812,700 460 373.8 6 .2
373.8 .
n
ad mm
aM T d N mm E N mm
KN m
Example 2.7
45
Calculate the nominal flexural strength of the beam if the yield stress of the steel is 420MPa and the strength of the concrete is 21MPa.
Example 2.7
46
Example 2.7
47
(a) Calculate the tensile force (T) and the compressive force (C); determine the area of concrete necessary to equilibrate T
2
'
2
'
2
2580 mm 420 1083.6
0.85
1083.6 360,705
0.85 0.85 21
(150)(150) ( 150)(450) 60,705
235
s y
c c
c
c
T A f MPa KN
C f A
C T
T EA mm
f MPa
a mm
a mm
Example 2.7
48
(c) Locate the NA and determine Mn.
Defining y as the distance from the top of the section to the centroid of the compression area,
8522500 75 38205 150
2149
60705
525 149 376
1083600 376 407 .n
y mm
d y mm
M N mm KN m