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2.diagram Interaksi P-M

The document discusses the design of reinforced concrete columns. It provides: 1) A diagram showing the interaction between axial load (P) and bending moment (M) for a reinforced concrete column. 2) Calculations of the nominal axial load capacity (Pn) and nominal bending moment capacity (Mn) of the column. 3) Design of shear reinforcement for the column, calculating the shear capacity (Vc) and required shear reinforcement (Vs). Shear studs with a spacing of 150 mm are determined to be sufficient.

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Deasy Naruth
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100 views8 pages

2.diagram Interaksi P-M

The document discusses the design of reinforced concrete columns. It provides: 1) A diagram showing the interaction between axial load (P) and bending moment (M) for a reinforced concrete column. 2) Calculations of the nominal axial load capacity (Pn) and nominal bending moment capacity (Mn) of the column. 3) Design of shear reinforcement for the column, calculating the shear capacity (Vc) and required shear reinforcement (Vs). Shear studs with a spacing of 150 mm are determined to be sufficient.

Uploaded by

Deasy Naruth
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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DIAGRAM INTERAKSI P - M

Data :
b = 500 mm Tulangan D =
h = 500 mm Luas 1-tul =
fc' = 25 MPa As,total =
fy = 390 MPa As = As' =
β1 = 0.85 =
φ = 0.65

GAMBARKAN DIAGRAM INTERAKSI P - M

Penyelesaian :
Tetapkan ds = d' = 60 mm
d = 440 mm

1.Penampang dalam keadaan seimbang :

cb =
600
⋅d
600+ fy
= 266.67 mm
ab = β 1⋅cb
= 226.67 mm

cb−d '
fs' = ⋅600
cb
= 465.00 MPa
fs' < fy ?
465.00 > 390.00 Digunakan fy !

Ts = As . fy
= 553104 N

Cs = As' . fy
= 553104 N

Cc = 0,85 . fc' . ab . b
= 2408333 N

Pb = Cc + Cs - Ts
Pb = 2408333 N
φ Pb = 1565417 N = 1565.417
Mb = Cc⋅ ( h2 − ab2 )+Cs⋅( h2 −d ' )+Ts⋅( h2 −ds)
Mb = 539318246.0317 Nmm
φMb = 350556859.9206 Nmm = 350.557

2. Gaya aksial sentris


PO = 0,85 f'c*Ag + As,t (fy - 0,85 f'c)
= 6358433.04 N
ΦPO = 4132981.47 N = 4132.98
Pn,mak = 0,8*Φ*PO
= 3306385.18 N = 3306.39

Pmin = 625000 N = 625.00

0,1 f'c b h = 625000 N = 625.00

2. Momen Lentur Murni

A = 0,85  fc 'b   1 = 9031


B = ( 600⋅As ' )− ( As⋅fy ) = 297825
C = −600⋅As '⋅d ' = -51055714

c = −B+ √ B 2−4⋅A⋅C = 60.49


2A
a = β 1⋅c = 51.41
c−d '
fs' = ⋅600 = 4.82
c

Cc = 0 , 85⋅fc '⋅a⋅b
= 546265 N
Cs = As' . fy
= 6839 N

Mn = Cc*(d-a/2) + Cs*(d-d')
= 228912648 Nmm
φMn = 0,9 Mn
= 206021383 Nmm = 206.021
φMn = 0,65 Mn
= 148793221 Nmm = 148.793
P-M

19 mm
283.64 mm2
10 D19
5 D19
1418.2143 mm2

nakan fy !

1565.417 kN
) ( h2 −ds)
−d ' +Ts⋅

350.557 kNm

kN

kN

kN

kN

9031
297825
-51055714

mm

mm

MPa

kNm
kNm
3. Tulangan Geser Kolom Persegi
Data :
Pu = 800 kN
Vu = 180 kN = 180000 N
fy = 240 MPa
fc' = 20 MPa
b = 500 mm
h = 500 mm
φ
β

= 0.6
1

Diambil ds = 55 mm

d = h−ds
= 445 mm

Vn = Vu
φ
= 300000 N

Pu 1
Vc =
( 1+
)( )
⋅ √ fc ' b⋅d
14⋅A gross 6
= 203748 N

φ Vc = 122249 N

Vu > φVc ?
180000 > 122249 Perlu tulangan geser !

Vs = Vn - Vc
= 96252 N

Digunakan sengkang P 10 Luas = 78.57 mm2


Av = 2 As
= 157.143 mm2

Av⋅fy⋅d
S perlu =
Vs
= 174.364 Dipakai P 10 - 150
Kontrol jarak sengkang :

Vs <
( 13 √ fc ' ) b⋅d ?

96252 < 331683 OK !

Digunakan :
S ma ks

S maks = d / 2
= 222.5 mm

Vs <
( 23 √ fc ' ) b⋅d ?

96252 < 663367 OK !

Digunakan :
S maks = d/4
= 111.25 mm

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