DRCS - Quick Reference (Unit 3)

Unit 3:

Shear in beams:

Forms of Shear Reinforcement:

1. Vertical Stirrups
2. Inclined Stirrups
3. Bent-up bars

Max. shear generally occur at a distance 'd' from the support. (d = effective depth of beam)

Types of Problems:

1. Design Shear Reinforcement (Vertical Stirrups only)

Procedure:

1. Given Data
2. Find Max. Shear Force, Vu (wl/2 for Simply Supported beam, wl for cantilever beam under udl)
3. Find Shear Stress Developed, Tv (tow v) = Vu / b d
4. Find Shear Strength of Concrete, Vc = Tc  bd 
5. Find Tc based on 1) % of Steel Reinforcement used 2) Grade of Concrete (Use IS Code book - Table 19)
6. Find % of Steel = (Ast x 100) / b d
7. If Vu > Vc, then Shear Reinforcement has to be provided
8. Find Shear Resistance offered by Vertical Stirrups, Vus = Vu - Vc
9. Assume 2 legged stirrups of 8mm dia bar
10. Find Spacing of Stirrups, Sv = (0.87fy Ast d) / Vus
11. Check for minimum spacing from the least of 1) 0.75d  2) 300mm 3) using Eqn.: Asv / b. Sv = 0.4 / 0.87fy
12. Draw cross section of beam and show shear reinforcement details

2. Design Shear Reinforcement (Vertical Stirrups + Bent-up Bars)

Procedure:

1. Given Data
2. Find Max. Shear Force, Vu (wl/2 for Simply Supported beam, wl for cantilever beam under udl)
3. Find Shear Stress Developed, Tv (tow v) = Vu / b d
4. Find Shear Strength of Concrete, Vc = Tc  bd 
5. Find based Tc based on 1) % of Steel Reinforcement used 2) Grade of Concrete (Use IS Code book - Table 19)
6. Find % of Steel = (Ast x 100) / b d
7. If Vu > Vc, then Shear Reinforcement has to be provided
8. Find Shear Resistance Required, Vus1 = Vu - Vc
9. Vus1 = Vus + Vub
10. Shear Resistance offered by Bent-up bars, Vub = 0.87 fy Asb (Sin α) (Limited to 1/2 of Vus1)
11. Assume α is 45⁰
12. Find Shear Resistance offered by Vertical Stirrups, Vus = Vus1 - Vub
13. Assume 2 legged stirrups of 8mm dia bar
14. Find Spacing of Stirrups, Sv = (0.87fy Ast d) / Vus
15. Check for minimum spacing from the least of 1) 0.75d  2) 300mm 3) using Eqn.: Asv / b. Sv = 0.4 / 0.87fy
16. Draw cross section of beam and show shear reinforcement details

Torsion: (Clause 41.1 of IS: 456-2000)

A beam subjected to torsion develops spiral cracks in structure

Types of Torsion:

1. Equilibrium Torsion - Ex. Sunshade, Cantilever Balcony
2. Compatibility Torsion - Ex. Ring Beams, Curved Beams

Equilibrium Torsion:

Torsion Resistance can be provided by providing additional Longitudinal Reinforcement and Shear Reinforcement

Types of Problem:

1. Design beam subjected to a Torsion

Procedure:

1. Given Data
2. Assume Clear Cover and find effective Cover (use 8mm dia for vertical stirrups)
3. Design Longitudinal Reinforcement, Mel = Mu + Mt
4. Find Mt = Tu [(1 + D/b) / 1.7] where Tu = Torsion Moment
5. Find Steel Reinforcement same as Singly Reinforced Beam
6. Design Max. Shear Force due to Torsion, Ve = Vu + (1.6 Tu / b)
7. Find Shear Reinforcement. Assume 8mm dia 2 legged stirrups
8. Spacing of Stirrups, Sv = Asv / [ Tu / (b1 d1 0.87fy) + Vu / (2.5 d1 0.87 fy)]
9. Find b1 = c/c of corner rods in the direction of width
10. Find d1 = c/c of corner rods in the direction of depth
11. Check for minimum spacing 1) Sv = Asv 0.87fy / (Tve - Tc) b 2) Sv = x1 3) Sv = (x1 + y1)/4  4) Sv = 300mm
12. x1 = c/c of stirrup width
13. y1 = c/c of stirrup depth
14. Provide Side Face Reinforcement of 12mm dia. bars 2 nos at the sides in the middle, if depth exceeds 450mm
15. Draw Cross Section of beam with Steel reinforcement details

Bond:

Bond is the binding force between steel and concrete that makes them act in an integrated manner.

Types:

• Anchorage Bond - Ex. Hooks, U - Bends
• Development Bond - reinforcement bars embedded into the concrete by the required length

Development Length of Bar, Ld = φ 0.87 fy / 4 τbd

Find τbd from IS Code for given grade of Concrete

Note:

• τbd for Tension can be increased by 60% for HYSD bars like Fe415 and Fe500 = 1.60 τbd
• τbd for Compression can be increased by 25% for plain bars like Fe250 = 1.25 τbd
• τbd for Compression can be increased by 60% for plain bars like Fe250 = = 1.60 x 1.25 τbd

Check for Bond:

In order to ensure bond between the positive moment reinforcement and the surrounding concrete, the dia. of tension reinforcement is limited to 

Ld < = (M1 / V) + Lo

Anchorage for Reinforcing bars:

• hooks
• bends - 1) 4 times dia. of bar for 45 degrees bend 2) 16 times of dia. of bar for U-bend