biaxial base plate2
TRANSCRIPT
8/13/2019 Biaxial Base Plate2
http://slidepdf.com/reader/full/biaxial-base-plate2 1/3
8/13/2019 Biaxial Base Plate2
http://slidepdf.com/reader/full/biaxial-base-plate2 2/3
8/13/2019 Biaxial Base Plate2
http://slidepdf.com/reader/full/biaxial-base-plate2 3/3
The Bolts are Safe in Shear
The Bolts are Safe in Tension
Interaction Check for Bolts
The Bolts are safe in Interaction Check
Summary
Length of plate : m B
Width of plate : m z
Pl. thickness (Tprovided) : cm
D
x x
B x D3
D x B3 z
Hence, Design Bearing Pressure for Plate = <
Plate is supported on two sides & two sides are free.
Refer Book ` Formula for Stress & Strain' By Roark & Young for value of b1
= x cms Hence b1 = For a/b =
Therefore,
b1 x q x b2
x x
= > = Revise thickness 2.8
9.871850.0
Trequired 3.20 cms Tprovided
T2 =
1.754 54.66 190.44=
60.0 Kg/cm2
= = 1850 Kg/cm2 Hence
T2
546.58 T/m2
Hence Safe
Plate Panel(axb) 13.7 13.8 1.754 0.99
54.66 Kg/cm2
+ 0.10 = 54.66 Kg/cm2
0.303 0.028 0.028 ) =Bearing Press.(qc) forMax. Comp.
= ( 39.13 + 11.47
= 0.008 m4
0.3025 m2
= 0.008 m4
provide 12 Nos of 39 mm dia Grade 8.8 Type Bolts
Ixx =12
Izz =12
2.8
A = =B x D
0.55
0.55
= 0.2615.48 23.23
+ (5.60
)= (0.26
)
DESIGN OF BASE PLATE
< 1.4Permisible Shear Force Permisible Tensile Force( Actual Tensile Force
max shear per
bolt
)( Actual Shear Force ) +
0.64 ton
1.600 12
=
< Shear Capacity of Bolt
1.600.
a b