• Tidak ada hasil yang ditemukan

lecture 9 pile cap design (cont.)

N/A
N/A
Protected

Academic year: 2023

Membagikan "lecture 9 pile cap design (cont.)"

Copied!
12
0
0

Teks penuh

(1)

CE 318

STRUCTURE ANALYSIS AND DESIGN II LAB

LECTURE 9

PILE CAP DESIGN (CONT.)

SEMESTER: SUMMER 2020

COURSE TEACHER: SAURAV BARUA CONTACT NO: +8801715334075

EMAIL: [email protected]

(2)

LECTURE PLAN

Net pile capacity

Effective depth and flexural moment

Punching shear and flexural shear check

Reinforcement calculation

Detailing

(3)

PILE CAP DESIGN

PC2 pile cap size = 9’-8” x 9’-8”

Load on pile cap = 467.67 kip

Net pile reaction, P

net

=

Load on pile cap nos. of pile

So, P

net

=

467.67

5

= 93.53 kip

(4)

1.5 d = 1.5 x 16 = 24” (here d= diameter of pile) 3√2 d = 68” (here d= diameter of pile)

dprovided = 3 x √(area of pile cap in sq. ft) +2 (thumb rule!!)

= 3 x √(9.67 x 9.67) +2 = 31”

Moment at face of the column

Mface = 2 x p x (68- 12/2) [assume, 12” wide column]

= 2 x 93.53 x 62 = 11598 k-in = 966.5 k-ft dreq = √[M/Øþfy b(1-0.59þfy/fc’)]

= √[11598/0.9x0.016x 60 x 12x (1-0.59x 0.016x60/3)]

= 37.13”

dreq > dprovided, so d ≈ 38” (effective depth of pile cap)

(5)

Punching shear check

Punching shear at d/2 distance from support. d/2 = 38/2 = 19”

Vp = 2 x p

= 2 x 93.53 = 187.06 kip

Punching shear stress,

v

p

= 𝑉

𝑝

𝑏

0

𝑑

Perimeter, b0 = 4 x (12+38) = 200”

vp = 187.06 x 1000/(200x38) = 24.61 psi 4Ø√fc’ = 4 x 0.85 x √3000 = 186.23 psi 4Ø√fc’ > vp ok

(6)

• Flexural shear check

Flexural shear at d distance from support d = 38”

Distance between column face to pile = (68/2 – 16/2-12/2) = 20”

flexural shear length = (38-20)= 18” of pile cap (16” diameter) So, Vm = 0 kip, no need to check for flexural shear

[Explain in class!!]

Flexural stress, vm = Vm/bd < 2Ø√fc’ = 2 x 0.85 x √3000 = 93.11 psi ok

(7)

• M = 966.5 k-ft/length = 966.5/9.67 = 99.94 k-ft [ pile cap size 9’-8” x 9’-8”]

As = M/Øfy (d-a/2) a = As fy/0.85 fc’ b

As = 99.94 x 12/[0.85 x 60 x (38-1/2)] ; assume a =1”

= 0.31 in2/ft

a = 0.31 x 60/(0.85 x 3 x 12) = 0.61”

As = 99.94 x 12/[0.85 x 60 x (38-0.61/2)] =0.62 in2/ft Asmin = (200/fy) bd = (200/6000) x 12 x 38

= 1.52 in2/ft

Astemp= 0.0018 bh = 0.0018 x 12 x (38+5)= 0.93 in2/ft (pile cap thickness h = 43”) Asmin will govern for both direction.

(8)

• Use Ø20mm rebar.

Spacing = 0.61x 12/1.52 = 4.5” c/c Provide Ø20mm @ 4.5”c/c in both ways

And thickness of pile cap = 38” + 3” +2” = 43” (Clear cover and space for pile reinforcement)

(9)

PILE CAP PLAN

a = Ø 20mm @ 4.5”c/c b = Ø 20mm @ 4.5”c/c

(10)

PILE CAP LONG SECTION

a = Ø 20mm @ 4.5”c/c b = Ø 20mm @ 4.5”c/c

(11)
(12)

Referensi

Dokumen terkait

If these measures fail, insulin should be instituted • Awareness, education and intensive counseling among the community and families will go a long way in prevention of T2DM in the

Contents 17 Indiana University at South Bend 1 Indiana University 18 Admission 1 Indiana University-Purdue University at 18 Professional Liability Insurance Indianapolis 18