Design Of Hinged Base
Data:H.E.B (300)
P = 1.84 t
H = 1.57 t
a = 30 cm
b = 30 cm
Fcu = 250 kg/cm2
1. Base Dimension
A = a +5 cm
A = 30 +5 = 35 cm
B1
B = b +5 cm
B
b
B = 30 +5 = 35 cm
a
A1 = A +10 cm
A
A1
A1 = 35 +10 = 45 cm
B1 = B +10 cm
B1 = 35 +10 = 45 cm
Area1 = 35*35 = 1225 cm2
Area2 = 45*45 = 2025 cm2
Area2
Area1
=
1.28
OK
2. Check Of concrete Capacity
Pu  C (0.67 * Fcu * Area1 )
Area2
Area1
Pu  0.6(0.67 * 0.25 *1225) *1.28
 157.58t
Pu  157.58  p  1.84  ok
3. Plate Thickness
pU
A1
 2  * M
* 2.5 * b  * 2.5
M P Fy * Z P
P
ZP
=
M P
b*tp
2
4
2
2.8 * 21 * t p
4
1.841225* 2.5 * 35 * 2.5 2  0.6 * 2.8 * t
t p = 0.3 cm
4. Design Of weld
2
p
Use Min. t p = 2 cm
Ou  Q  0.6 * N 
QU  1.84  0.6 *1.57
= 2.78 t
Ruw,q  Q
Lw
= 2.78
2 * 35  2 *1.1
 0.7 * S * 0.4 * Fu 
= 0.3 cm
Take  Sweld  Smin .  0.5cm
5. Design Of Bolts
For bolts Grade (10.9) , M=24
RUV =1.84 / 2
= 0.92 t
R shear  0.6(0.6Fub ) As * n
0.92  0.6(0.6 * 6) As *1
AS
= 0.43
Use 2 M 20
R bearing  br * d * t min * ( * Fu )
R bearing  0.7 * 2.0 * 2 * (1.2 * 4.4)
= 14.7 t
Rmin  14.7ton
N

Qu
Rmin
1.84
 0.12
14.7
Take  N  2