18
9.1 ksi f ksi f y c 60 4 ' = = Example 1 – Design of Cap Beam
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
