1. Lecture 21 – Splices and Shear
February 5, 2003
CVEN 444

2. Lecture Goals
Spice
Shear
Shear Design

3. Bar Splices Why do we need bar splices? -- for long spans
Types of Splices
1. Butted &Welded
2. Mechanical Connectors
3. Lay Splices
Must develop 125% of yield strength

4. Tension Lap Splices Why do we need bar splices? -- for long spans
Types of Splices
1. Contact Splice
2. Non Contact Spice (distance 6” and 1/5 splice length)
Splice length is the distance the two bars are overlapped.

5. Types of Splices Class A Spice (ACI 12.15.2)
When over entire splice length.
and 1/2 or less of total reinforcement is spliced win the req’d lay length.

6. Types of Splices Class B Spice (ACI 12.15.2)
All tension lay splices not meeting requirements of Class A Splices

7. Tension Lap Splice (ACI 12.15)
where As (req’d) = determined for bending
ld = development length for bars (not allowed to use excess reinforcement modification factor)
ld must be greater than or equal to 12 in.

8. Tension Lap Splice (ACI 12.15)
Lap Spices shall not be used for bars larger than No. 11. (ACI )
Lap Spices should be placed in away from regions of high tensile stresses -locate near points of inflection (ACI )

9. Compression Lap Splice (ACI 12.16.1)
Lap, req’d = fy db for fy < psi Lap, req’d = (0.0009fy -24) db for fy > psi Lap, req’d in
For fc psi, required lap splice shall be multiply by (4/3) (ACI )

10. Compression Lap Splice (ACI 12.17)
In tied column splices with effective tie area throughout splice length hs factor = 0.83
In spiral column splices, factor = But final splice length in.

11. Example – Splice Tension
Calculate the lap-splice length for 6 #8 tension bottom bars in two rows with clear spacing 2.5 in. and a clear cover, 1.5 in., for the following cases
When 3 bars are spliced and As(provided) /As(required) >2
When 4 bars are spliced and As(provided) /As(required) < 2
When all bars are spliced at the same location fc= 5 ksi and fy = 60 ksi a. b. c.

12. Example – Splice Tension
For #8 bars, db =1.0 in. and a = ?b =? g = ?l =?

13. Example – Splice Tension
The As(provided) /As(required) > 2, class ? splice applies;
The As(provided) /As(required) < 2, class ? splice applies;

14. Example – Splice Compression
Calculate the lap splice length for a # 10 compression bar in tied column when fc= 5 ksi and when a) fy = 60 ksi and b) fy = 80 ksi

15. Example – Splice Compression
For #10 bars, db =? in.
Check ls > db fy

16. Example – Splice Compression
For #10 bars, db =? in. The ld = 2? in.
Check ls > ( fy –24) db
So use ls = ? in.

17. Shear Design

18. Uncracked Elastic Beam Behavior
Look at the shear and bending moment diagrams. The acting shear stress distribution on the beam.

19. Uncracked Elastic Beam Behavior
The acting stresses distributed across the cross-section.
The shear stress acting on the rectangular beam.

20. Uncracked Elastic Beam Behavior
The equation of the shear stress for a rectangular beam is given as:
Note: The maximum 1st moment occurs at the neutral axis (NA).

21. Uncracked Elastic Beam Behavior
The ideal shear stress distribution can be described as:

22. Uncracked Elastic Beam Behavior
A realistic description of the shear distribution is shown as:

23. Uncracked Elastic Beam Behavior
The shear stress acting along the beam can be described with a stress block:
Using Mohr’s circle, the stress block can be manipulated to find the maximum shear and the crack formation.

24. Inclined Cracking in Reinforced Concrete Beams
Typical Crack Patterns for a deep beam

25. Inclined Cracking in Reinforced Concrete Beams
Flexural-shear crack - Starts out as a flexural crack and propagates due to shear stress.
Flexural cracks in beams are vertical (perpendicular to the tension face).

26. Inclined Cracking in Reinforced Concrete Beams
For deep beam the cracks are given as:
The shear cracks Inclined (diagonal) intercept crack with longitudinal bars plus vertical or inclined reinforcement.

27. Inclined Cracking in Reinforced Concrete Beams
For deep beam the cracks are given as:
The shear cracks fail due two modes:
- shear-tension failure shear-compression failure

28. Shear Strength of RC Beams without Web Reinforcement
Total Resistance = vcz + vay + vd (when no stirrups are used)
vcz - shear in compression zone
va - Aggregate Interlock forces
vd = Dowel action from longitudinal bars
Note: vcz increases from (V/bd) to (V/by) as crack forms.

29. Strength of Concrete in Shear (No Shear Reinforcement)
(1) Tensile Strength of concrete affect inclined cracking load

30. Strength of Concrete in Shear (No Shear Reinforcement)
(2) Longitudinal Reinforcement Ratio, rw

31. Strength of Concrete in Shear (No Shear Reinforcement)
(3) Shear span to depth ratio, a/d (M/(Vd))
Deep shear spans more detail design required
Ratio has little effect

32. Strength of Concrete in Shear (No Shear Reinforcement)
(4) Size of Beam Increase Depth Reduced shear stress at inclined cracking

33. Strength of Concrete in Shear (No Shear Reinforcement)
(5) Axial Forces Axial tension Decreases inclined cracking load - Axial Compression Increases inclined cracking load (Delays flexural cracking)

34. Function and Strength of Web Reinforcement
Web Reinforcement is provided to ensure that the full flexural capacity can be developed. (desired a flexural failure mode - shear failure is brittle)
- Acts as “clamps” to keep shear cracks from widening
Function:

35. Function and Strength of Web Reinforcement
Uncracked Beam Shear is resisted uncracked concrete.
Flexural Cracking Shear is resisted by vcz, vay, vd

36. Function and Strength of Web Reinforcement
Flexural Cracking Shear is resisted by vcz, vay, vd and vs
Vs increases as cracks widen until yielding of stirrups then stirrups provide constant resistance.

37. Designing to Resist Shear
Shear Strength (ACI 318 Sec 11.1)

38. Designing to Resist Shear
Shear Strength (ACI 318 Sec 11.1)
Nominal shear resistance provided by concrete
Nominal shear provided by the shear reinforcement

39. Shear Strength Provided by Concrete
Bending only
Simple formula
More detailed
Note:
Eqn [11.3]
Eqn [11.5]

40. Shear Strength Provided by Concrete
Bending and Axial Compression
Nu is positive for compression and Nu/Ag are in psi.
Simple formula
Eqn [11.4]
Eqn [11.7]

41. Typical Shear Reinforcement
Stirrup - perpendicular to axis of members (minimum labor - more material)
ACI Eqn 11-15

42. Typical Shear Reinforcement
Bent Bars (more labor - minimum material) see req’d in
ACI

43. Stirrup Anchorage Requirements
Vs based on assumption stirrups yield
Stirrups must be well anchored.

44. Stirrup Anchorage Requirements
Refer to Sec of ACI 318 for development of web reinforcement. Requirements:
each bend must enclose a long bar
# 5 and smaller can use standard hooks 90o,135o, 180o
#6, #7,#8( fy = 40 ksi )
#6, #7,#8 ( fy > 40 ksi ) standard hook plus a minimum embedment

45. Stirrup Anchorage Requirements
Also sec requirement for minimum stirrups in beams with compression reinforcement, beams subject to stress reversals, or beams subject to torsion

46. Design Procedure for Shear
(1) Calculate Vu
(2) Calculate fVc Eqn 11-3 or 11-5 (no axial force)
(3) Check
If yes, add web reinforcement (go to 4)
If no, done.

47. Design Procedure for Shear
(4)
Provide minimum shear reinforcement
Also:
(Done)

48. Design Procedure for Shear
(5)
Check:

49. Design Procedure for Shear
(6)
Solve for required stirrup spacing(strength) Assume # 3, #4, or #5 stirrups
from 11-15

50. Design Procedure for Shear
(7) Check minimum steel requirement (eqn 11-13)

51. Design Procedure for Shear
(8) Check maximum spacing requirement (ACI )

52. Design Procedure for Shear
(9) Use smallest spacing from steps 6,7,8
Note: A practical limit to minimum stirrup spacing is 4 inches.

53. Location of Maximum Shear for Beam Design
Non-pre-stressed members:
Sections located less than a distance d from face of support may be designed for same shear, Vu, as the computed at a distance d.
Compression fan carries load directly into support.

54. Location of Maximum Shear for Beam Design
The support reaction introduces compression into the end regions of the member.
No concentrated load occurs with in d from face of support . 1. 2.
When:

55. Location of Maximum Shear for Beam Design
Compression from support at bottom of beam tends to close crack at support

56. Homework
Determine the development length required for the bars shown . fc =4-ksi and fy = 60-ksi. Check the anchorage in the column. If it is not satisfactory, design an anchorage using a 180o hook and check adequacy.

57. Homework
Considering the anchorage of the beam bars into a column, determine the largest bar that can be used with out a hook. fc = 3-ksi and fy= 40ksi

58. Homework
A simple supported uniformly loaded beam carries a total factored design load of 4.8 k/ft (including self-weight) on a clear span of 34 ft. fc =3 ksi and fy=40 ksi. Assume that the supports are 12 in wide and assume that the bars are available in 30 ft lengths.
Design a rectangular beam
Determine bar cutoffs.
Locate splices and determine the lap length.