Strut and Tie Design Example

9.1

ksif

ksif

y

c

60

4'

=

=

Example 1 – Design of Cap Beam

9.2Design Steps

1. Visualize flow of stresses and Sketch an idealized strut-and-tie model

2. Check size of bearing – nodal zone stresses3. Select area of ties4. Check strength of struts5. Provide adequate anchorage for the ties6. Provide crack control reinforcement7. Sketch required reinforcement

9.3

Step 1 - Draw Idealized Truss ModelPier Cap Elevation

9.4Step 1 - Draw Idealized Truss Model

9.5Step 1 – Solve for Member Forces

9.6Steps 2 thru 5 – Check Strength

2. Size of Bearing

3. Tension Tie

4. Compression Strut

5. Anchorage

9.7Step 2 – Check Size of Bearings

2'c

u in. 142470.065.0

259f0.65

Prequired area bearing =

××=

φ=

'85.0 cf of stress limiting – node CCC – D Node φ'75.0 cf of stress limiting – node CCT A – Node φ

'65.0 cf of stress limiting –node CTT – B Node

9.8

Step 3 – Choose Tension Tie Reinforcement

2in. 46.5609.0

295=

×==

y

ust f

PAφ

Use 6 No. 9 bars 2s in. 0.6A =

a) Top Reinforcement over Column, Tie AB

9.9

Step 3 – Choose Tension Tie Reinforcement

Use 12 No. 9 bars 2in. 0.12=sA

b) Bottom Reinforcement at Midspan

2in. 20.11609.0

605=

×==

y

ust f

PAφ

9.10

Step 3 – Choose Tension Tie Reinforcement

Provide No. 5 double-legged stirrups at 12 in

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

45.46031.029.0

149fA

Pn

yst

u =×××

=φ

=

in. 5.1345.4

60s =≤

9.11Step 4 – Check Capacity of Struts

• Strut FB is most critical

• fcu controlled by tensile strain in tie at smallest angle to strut

3

sst

us 10695.1

000,290.6295

EAP −×=

×==ε

9.12Step 4 – Check Capacity of Struts

671 kips

295 kips

285kips

217kips

= 1.695×10-3

εs = 0.848×10-3

18" bearing

8"

ε ≈ 0

18 sin 40.3° + 8 cos 40.3° = 17.7”

εs

40.3o

εs = 1.657×10-3

149 kips

49.7o

( ) 33s 10848.02/010695.1 −− ×=+×=ε

( ) ( ) 30233s

2ss1 1081.43.40cot002.010848.010848.0cot002.0 −−− ×=+×+×=α+ε+ε=ε

9.13Step 4 – Check Capacity of Struts

671 kips

295 kips

285kips

217kips

= 1.695×10-3

εs = 0.848×10-3

18" bearing

8"

ε ≈ 0

18 sin 40.3° + 8 cos 40.3° = 17.7”

εs

40.3o

εs = 1.657×10-3

149 kips

49.7o

ksi 3.4040.85ksi 47.21081.41708.0

4f85.01708.0

ff 3

'c

1

'c

cu =×≤=××+

=≤ε+

=−

kips 1312307.1747.2AfP cscun =××==required kips 671kips 918131270.0PP nr ≥=×=φ=

40.3O

9.14

Step 5 – Check Anchorage of Tension Tie

43 in.

• 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

9.15Step 5 – Check Anchorage of Tension Tie

• Check nodal zone stress

ksi 024.13642

295fc =××

=

• Limiting nodal zone stress (5.6.3.6) is:

ksi 1.2470.075.0f75.0f 'cc =××=φ=

9.16Step 6 – Provide Crack Control Reinforcement

• D-region (region near discontinuity)

• Between nodes A & B• 0.003 of Gross Area

• Section §5.6.3.6

2s in. 30.13612003.0A =××=

• Provide 4 No. 5 bars horizontal (1.24 in2)

• Provide 4 legs of No. 5 stirrups

9.17Step 6 – Provide Crack Control Reinforcement

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

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

2

y

v'cv in. 46.0

60123640316.0

fsb

f0316.0A =×

××==

9.18Step 7 – Sketch the Required Reinforcement

D - region B - region

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

2 legged No.5 stirrups at 12"

4 – No.5 2 – No.5 12 – No.9

4 – No.5 typ. 2 – No.5 typ. each face

6 – No.9 top

4 legs of No.5 closed stirrups @ 12"

12 – No.9 bot 12 – No.9 bot

2 No.5@12"

2 – No.9 top