Baseplate Cirrcle 2018 - HUTATI LAP
Baseplate Cirrcle 2018 - HUTATI LAP
Baseplate Cirrcle 2018 - HUTATI LAP
1. Input data
Axial load P= 889.64 kN
moment load M= 2259.69 kN.m
Eccentricity e = M/N = 2.54 m
Outside diameter of the steel column pipe Do = 1066.80 mm
Ratio of concrete to plate areas A2/A1 = 1.50
Allowable stresses for the base plate Fy = 344.74 MPa
Allowable stresses for the anchor bolts Fy,b = 303.37 MPa
The concrete strength f'c = 34.47 MPa
2. Calculation design
2.1. Base plate thickness calculated
The maximum design bearing stress is
F_p=0.35× 〖 f′ 〗 _c×√(A2/A1)
= 14.78 MPa
= 66.63 dgree
The chord of the segment is: B=2×sinα×N/2 = 1398.98 mm
and area of the segment (compressive bearing area)
A_seg=0.0087266×(N/2)^2×2α−B(N/2−A)/2 = 463808.09 mm2
Determine the c.g. of the segment:
C=N/2 ((2×sin^3α)/3(α−sinα 〖 .cos 〗α ) −cosα )
If α ≥ π/4 = 0.785 rad; = 189.67 mm
C=0.2×N/2×α^2 (1−0.0619α^2+0.0027α^4 )
If α < π/4 = 0.785 rad; =
Where: C_distance from the c.g. of compressive stress bearing area to the zero stress section.
Determine the values of both the left and right sides:
On the left of the equation: P(e+A^′ ) = 2657.59 kN.m
F_p×C/A×A_seg [N/2−(A−C)+A′]
Second member of equation: = 2655.76 kN.m
Since: the difference between the left and right sides of Eq. only 0.07%. Eq. is satisfied
Determine T (sum of the anchor bolt forces in the half of the circle)
T=F_p×C/A×A_seg−P
= 1938.08 kN
Determine Tmax (where Tmax = T1_outmost anchor bolt force)
T_max=T/(1+2/y_1 ∑24_(i=2)^n▒y_1 )
= 255.15 kN
= 45.57 dgree
The chord of the segment is B1 = 1088.35 mm
and the area of the segment is Aseg1 = 171577.18 mm2
Since:
If α1 ≥ π/4 = 0.785 rad; C1 = 92.73 mm
If α1 < π/4 = 0.785 rad; C1 =
t_(p,req)=√((6M_pl1)/F_b )
Determine the base plate thickness: = 59.43 mm
Where:
Moment for a 1 mm wide strip at the critical section, i.e., the total moment
at the critical section divided by the chord at the critical section:
M_pl1=[(A−0.5(N−D_o )+C1)/A]((F_p×A_seg1×C1)/B1)
= 0.15 kN.m
= -64.97 kN
α1=arccos((Do/2)/(N/2))
Check tensile force on bolt:
= 111.50 kN - Pass
Assuming the stiffeners are provided:
If the stiffeners are provided, the critical section is located at 109.5 mm from the centerline of the
base plate. Similar to the previous calculation, consider the circular segment at the critical section
and calculate the section properties of the segment as follows:
t_(p,req)=√((6M_pl1)/F_b )
= 1.063 rad
= 60.88 dgree
The chord of the segment is B1 = 393.11 mm
M_pl1=[(A−0.5(N−D_o
and the area of the segment is )+C1)/A]((F_p×A_seg1×C1)/B1)
A = 32267.09 mm2
F_b=0.75F_y
seg1
Since:
If α1 ≥ π/4 = 0.785 rad; C1 = 47.40 mm
If α1 < π/4 = 0.785 rad; C1 =
R_c=F_p×C/A×A_seg
Where:
Moment for a 1 mm wide strip at the critical section, i.e., the total moment
at the critical section divided by the chord at the critical section:
= 0.03 kN.m