1. Lecture 5 - Flexure
June 11, 2003
CVEN 444

2. Lecture Goals Rectangular Beams Safety factors Loading and Resistance
Balanced Beams

3. Flexural Stress
The compressive zone is modeled with a equivalent stress block.

4. Flexural Stress
The equivalent rectangular concrete stress distribution has what is known as a b1 coefficient is proportion of average stress distribution covers.

5. Flexural Stress Requirements for analysis of reinforced concrete beams
[1] Stress-Strain Compatibility – Stress at a point in member must correspond to strain at a point.
[2] Equilibrium – Internal forces balances with external forces

6. Flexural Stress Example of rectangular reinforced concrete beam.
(1) Setup equilibrium.

7. Flexural Stress Example of rectangular reinforced concrete beam.
(2) Find flexural capacity.

8. Flexural Stress Example of rectangular reinforced concrete beam.
(2) Find flexural capacity.

9. Flexural Stress Example of rectangular reinforced concrete beam.
(3) Need to confirm es > ey

10. Flexural Stress –Rectangular Example
Example of rectangular reinforced concrete beam.
Given a rectangular beam
fc = 4000 psi
fy = 60 ksi (4 #7 bars)
b = 12 in. d = 15.5 in. h= 18 in.
Find the neutral axis.
Find the moment capacity of the beam.

11. Flexural Stress –Rectangular Example
Determine the area of steel, #7 bar has 0.6 in2.
The b value is b1 = 0.85 because the concrete has a fc =4000 psi.

12. Flexural Stress –Rectangular Example
From equilibrium (assume the steel has yielded)
The neutral axis is

13. Flexural Stress –Rectangular Example
Check to see whether or not the steel has yielded.
Check the strain in the steel
Steel yielded!

14. Flexural Stress –Rectangular Example
Compute moment capacity of the beam.

15. Flexural Stress – Non-Rectangular Example
For a non-rectangular beam
For the given beam with concrete rated at fc = 6 ksi and the steel is rated at fs = 60 ksi. d = 12.5 in.
(a) Determine the area of the steel for a balanced system for shown area of concrete.
(b) Determine the moment capacity of the beam. Mn
(c) Determine the NA.

16. Flexural Stress – Non-Rectangular Example
For a non-rectangular beam
The area of the concrete section is
The force due to concrete forces.

17. Flexural Stress – Non-Rectangular Example
Using equilibrium, the area of the steel can be found

18. Flexural Stress – Non-Rectangular Example
Find the center of the area of concrete area

19. Flexural Stress – Non-Rectangular Example
The moment capacity of the beam is

20. Flexural Stress – Non-Rectangular Example
Compute the b1 value

21. Flexural Stress – Non-Rectangular Example
Find the neutral axis

22. Safety Provisions
Structures and structural members must always be designed to carry some reserve load above what is expected under normal use.

23. Safety Provisions
There are three main reasons why some sort of safety factor are necessary in structural design.
[1] Consequences of failure.
[2] Variability in loading.
[3] Variability in resistance.

24. Consequences of Failure
A number of subjective factors must be considered in determining an acceptable level of safety.
Potential loss of life.
Cost of clearing the debris and replacement of the structure and its contents.
Cost to society.
Type of failure warning of failure, existence of alternative load paths.

25. Variability in Loading
Frequency distribution of sustained component of live loads in offices.

26. Variability in Resistance
Variability of the strengths of concrete and reinforcement.
Differences between the as-built dimensions and those found in structural drawings.
Effects of simplification made in the derivation of the members resistance.

27. Variability in Resistance
Comparison of measured and computed failure moments based on all data for reinforced concrete beams with fc > 2000 psi.

28. Margin of Safety
The distributions of the resistance and the loading are used to get a probability of failure of the structure.

29. Margin of Safety The term Y = R - S
is called the safety margin. The probability of failure is defined as:
and the safety index is

30. Loading SPECIFICATIONS
Cities in the U.S. generally base their building code on one of the three model codes:
Uniform Building Code
Basic Building Code (BOCA)
Standard Building Code

31. Loading
These codes have been consolidated in the 2000 International Building Code.
Loadings in these codes are mainly based on ASCE Minimum Design Loads for Buildings and Other Structures has been updated to ASCE 7-02.

32. Loading
The loading variations are taken into consideration by using a series of “load factors” to determine the ultimate load, U.

33. Loading
The equations come from ACI code 9.2 on loading (4.6 in your book),
D – Dead Load W – Wind Load
L – Live Load Lr – Roof Load
F – Fluid Pressure R – Rain Load
E – Earthquake Load T – Temperature Load
S – Snow Load H – Soil Load

34. Loading
The most general equation for the ultimate load, U (Mu) that you will see is going to be:

35. Resistance
The load factors will generate the ultimate load, which is used in the design and analysis of the structural member.
Mu – Ultimate Moment
Mn – Nominal Moment
f – Strength Reduction Factor

36. Resistance
The strength reduction factor, f, varies from member to member depending whether it is in tension or compression or the type of member. The code has been setup to determine the reduction.

37. Three possibilities in Inelastic Behavior
Compression Failure - (over-reinforced beam)
Tension Failure - (under-reinforced beam)
Balanced Failure - (balanced reinforcement)

38. Inelastic Behavior Compression Failure
The concrete will crush before the steel yields. This is a sudden failure.
The beam is known as an over-reinforced beam.

39. Inelastic Behavior Tension Failure
The reinforcement yields before the concrete crushes. The concrete crushes is a secondary compression failure.
The beam is known as an under-reinforced beam.

40. Inelastic Behavior Balanced Failure
The concrete crushes and the steel yields simultaneously.
The beam is known as an balanced-reinforced beam.

41. Inelastic Behavior Which type of failure is the most desirable?
The under-reinforced beam is the most desirable.
fs = fy
es >> ey
You want ductility
system deflects and still carries load.

42. Balanced Reinforcement Ratio, rbal
rbal = unique r value to get simultaneous ec = & es = ey
Use similar triangles:

43. Balanced Reinforcement Ratio, rbal
The equation can be rewritten to find cb

44. Nominal Moment Equation
The equation can be rewritten in the form:

45. Nominal Moment Equation
The equation can be rewritten in the form:
Use the ratio r = b/d and r

46. Nominal Moment Equation
Use w =r fy/fc and
Use the ratio r = b/d and R

47. Strain Limits Method for Analysis
The strength reduction factor, f, will come into the calculation of the strength of the beam.

48. Limitations on Reinforcement Ratio, r
The selection of the steel will be determined by the
Lower Limit on r ACI
ACI Eqn. (10-3)
fc & fy are in psi

49. Limitations on Reinforcement Ratio, r
Lower Limit on r ACI
Lower limit used to avoid “Piano Wire” beams.
Very small As ( Mn < Mcr )
es is huge (large deflections)
when beam cracks ( Mn > Mcr ) beam fails right away because Mn < Mcr

50. Additional Requirements for Lower Limit on r
If As (provided) 4/3 As (required) based on analysis, then As (min) is not required.
for As (provided)
See ACI

51. Additional Requirements for Lower Limit on r
Temperature and Shrinkage reinforcement in structural slabs and footings (ACI 7.12) place perpendicular to direction of flexural reinforcement.
GR 40 or GR 50 Bars: As (T&S) = Ag
GR 60 or Welded Wire Fabric (WWF):
As (T&S) = Ag
Ag - Gross area of the concrete

52. Example Given: fc = 3 ksi & fy = 40 ksi and As = 4 in2 Determine: (1)
(2)
Determine if the beam will satisfy ACI code.
If fc = 6 ksi?

53. Example Given: fc = 3 ksi & fy = 40 ksi and As = 4 in2
The minimum steel ratio is

54. Example Given: fc = 3 ksi & fy = 40 ksi and As =4 in2
The neutral axis is

55. Example The strain in the steel is
There for the beam is in the compression zone and f would be 0.65, however c/d ratio is greater than so the beam will need to be redesigned.

56. Example
c/d=0.615

57. Example Given: fc = 6 ksi & fy = 40 ksi and As =4 in2
The minimum steel ratio is

58. Example Given: fc = 6 ksi & fy = 40 ksi and As =4 in2
The neutral axis is at

59. Example The strain in the steel will be
There for the beam is in the tension zone and f will be 0.9.

60. Example
c/d=0.349