Strut and Tie Design Example

9.1

Example 1 – Design of Cap Beam

f c' = 4ksi f y = 60ksi

Design Steps

1. Visualize flow of stresses and Sketch an idealized strut-and-tie model 2. Check size of bearing – nodal zone stresses 3. Select area of ties 4. Check strength of struts 5. Provide adequate anchorage for the ties 6. Provide crack control reinforcement 7. Sketch required reinforcement

9.2

9.3

Step 1 - Draw Idealized Truss Model Pier Cap Elevation

9.4

Step 1 - Draw Idealized Truss Model

9.5

Step 1 – Solve for Member Forces

Steps 2 thru 5 – Check Strength 5. Anchorage

3. Tension Tie 2. Size of Bearing

4. Compression Strut

9.6

Step 2 – Check Size of Bearings

Node D – CCC node – limiting stress of 0.85φ f c' Node A – CCT node – limiting stress of 0.75φf c' Node B – CTT node – limiting stress of

Pu

' 0.65 f c

259 2 bearing area required = = = 142 in. 0.65φf c' 0.65 × 0.70 × 4

9.7

9.8

Step 3 – Choose Tension Tie Reinforcement

a) Top Reinforcement over Column, Tie AB

Pu 295 2 = = 5.46 in. Ast = φ f y 0.9 × 60 Use 6 No. 9 bars

A s = 6.0 in.

2

9.9

Step 3 – Choose Tension Tie Reinforcement

b) Bottom Reinforcement at Midspan

Pu 605 2 Ast = = = 11.20 in. φ f y 0.9 × 60 Use 12 No. 9 bars

As = 12.0 in.

2

9.10

Step 3 – Choose Tension Tie Reinforcement

c) Stirrups, Ties BG & CH Try 2-legged No. 5 Stirrups

Pu 149 n= = = 4.45 φA st f y 0.9 × 2 × 0.31 × 60

60 s≤ = 13.5 in. 4.45 Provide No. 5 double-legged stirrups at 12 in

Step 4 – Check Capacity of Struts

• Strut FB is most critical • fcu controlled by tensile strain in tie at smallest angle to strut

Pu 295 εs = = = 1.695 × 10 −3 A st E s 6.0 × 29,000

9.11

9.12

Step 4 – Check Capacity of Struts 285 kips 18" bearing

εs = 1.695×10-3

ε≈0

295 kips

8" 40.3

217 kips

o

εs = 0.848×10-3 49.7

εs = 1.657×10-3

o

149 kips

671 kips

18 sin 40.3° + 8 cos 40.3° = 17.7”

(

)

ε s = 1.695 × 10 −3 + 0 / 2 = 0.848 × 10 −3

(

)

ε1 = ε s + (ε s + 0.002 ) cot 2 α s = 0.848 × 10 −3 + 0.848 × 10 −3 + 0.002 cot 2 40.3 0 = 4.81 × 10 −3

Step 4 – Check Capacity of Struts

9.13

285 kips 18" bearing

εs = 1.695×10-3

ε≈0 40.3

295 kips 40.3

O

8"

217 kips

o

εs = 0.848×10-3 49.7 671 kips

εs = 1.657×10-3

o

149 kips 18 sin 40.3° + 8 cos 40.3° = 17.7”

f cu

f c' 4 = ≤ 0.85f c' = = 2.47 ksi ≤ 0.85 × 4 = 3.40 ksi − 3 0.8 + 170ε1 0.8 + 170× 4.81 × 10

Pn = f cu A cs = 2.47 × 17.7 × 30 = 1312 kips Pr = φPn = 0.70 × 1312 = 918 kips ≥ 671 kips required

9.14

Step 5 – Check Anchorage of Tension Tie

• Embedment length for No. 9 bars = 36 + 9 – 2 in. cover = 43 in. • Development length for No. 9 bars including top bar effect = 48 in. • Provide hooked bars

43 in.

Step 5 – Check Anchorage of Tension Tie

• Check nodal zone stress 295 fc = = 1.024 ksi 2 × 4 × 36

• Limiting nodal zone stress (5.6.3.6) is: f c = 0.75φf c' = 0.75 × 0.70 × 4 = 2.1 ksi

9.15

Step 6 – Provide Crack Control Reinforcement

• • •

D-region (region near discontinuity) Between nodes A & B 0.003 of Gross Area

•

Section §5.6.3.6

A s = 0.003 × 12 × 36 = 1.30 in. 2 • •

Provide 4 No. 5 bars horizontal (1.24 in2) Provide 4 legs of No. 5 stirrups

9.16

Step 6 – Provide Crack Control Reinforcement

• • •

A v = 0.0316 f c' • •

B-region (flexural region) Between nodes B & D Minimum Av per §5.8.2.5

b vs 36 × 12 = 0.0316 × 4 × = 0.46 in. 2 fy 60

Provided 2-legged No. 5 stirrups at 12 in. OK

9.17

Step 7 – Sketch the Required Reinforcement D - region

B - region

4 legged No.5 6 – No.9 stirrups at 12"

2 legged No.5 stirrups at 12"

4 – No.5

2 – No.5

6 – No.9 top 4 legs of No.5 closed stirrups @ 12" 4 – No.5 typ.

12 – No.9

2 – No.9 top 2 No.5@12" 2 – No.5 typ. each face

12 – No.9 bot

12 – No.9 bot

9.18