Design Example 3 Rectangular Silo Design a single rectangular concrete silo for storing peas. The bottom is a symmetrical pyramidal Hopper. The silo walls rest on the Hopper base which is supported by four 2
2
columns. The Roof load ( DL = 150 kg/m and LL= 100 kg/m ). Use f c' = 350 kg / cm 2 , f y = 4200 kg / cm 2
b=6m 30m
m 4 = a
An Above Hopper
5m
b=6m
3m 7m
a=6m
Openning0.5x0.5m
φ50cm
Solution
For Peas γ = 800 kg / m 3 φ = 25o µ ' = 0.296
ENGC6353
Dr. Mohammed Arafa
Page 1
Assume angle of response ρ =φ =25 hs = 3 tan 25 = 1.4 m
⇒
2 3
hs ≅ 1.0 m
k = 1 − sin 25 = 0.577 R a =
a
=
4
= 1.0m 4 4 a' R b = = 1.0m 4 2× 4×6 a'= = 4.8 4+6
Rb =
a'
4
= 1.2 m
Overpressure Factor C d 25cm
H / D = 40 /10 = 4
upper H1 c d = 1.5 lower 2/3 H
c d = 1.85
Hooper c d = 1.5
At the bottom of the silos At the bottom of the silos Y = 30 -1.0 = 29.0m q=
γ R
1 − e −( µ ' kY µ ' k
R)
p = kq For short wall ( R = 1.0)
2 q = 4.65 t/m
p = kq = 0.577 × 4.65 = 2.7 t/m For long wall ( R = 1.2)
q = 5.53t/m
2
2
p = kq = 0.577 × 5.53 = 3.2 t/m
2
Vertical Loads Due to Friction Friction
V = ( γ Y − q ) R
Short Wall
V = ( 0.8 × 30 − 4.65 ) ×1.0 = 19.35 ton
Long Wall
V = ( 0.8 × 30 − 5.53 ) ×1.2 = 22.16ton
Wall tension and bending moment
Short Wall
F a ,u = 1.7 (1.85 × 3.2 ) × 6 2 = 30.2ton/m
Long Wall
F b ,u = 1.7 (1.85 × 2.7 ) × 4 2 = 17.0ton/m
Frame action analysis using moment distribution Analysis
Assume wall thickness h=30cm ENGC6353
Dr. Mohammed Arafa
Page 2
The moment distribution is computed for an idealized rectangular frame 6.3 by 4.4 m Using symmetry K a =
Short Wall
K b =
Long Wall DFa =
0.465
0.465 + 0.317 DF b ≅ 0.4
2 I La
2 I Lb
2I
=
4.3
=0.465I 2
=
2I 6.3
m / t 7 . 2
=0.317I
4.3m 2
3.2 t/m
≅ 0.6 6.3m 7.9 t.m
Short Wall DF
Long Wall
0.6
0.4
FEM
4.16
-10.6
Balancing
3.86
2.58
FINAL
8.0
-8.0
-8.0 -8.0 5 . 1 -
Assume fillit (hunch) at the corner =25cm Negative moment will be calculated at the face of the hunch M b,-ve = 8.0 + 3.2 × 0.4 2 / 2 −10.1 × 0.4 = 4.2 t . m M a,-ve = 8.0 + 2.7 × 0.4 / 2 − 5.8 × 0.4 = 5.9 t . m 2
Check for thickness T 6M 2 + 2 ≤ f r = 2 f c ' = 2 350 = 37.4 kg / cm t ,b = bt bt For long Wall
25cm
5 17 ×103 6 ( 4.2 ×10 ) = 11kg / cm 2 < f r = + 2 1.7 ×1.85 (30 )(100 ) ( 30 ) (100 )
1
t ,b
For short Wall
30.2 ×10 3 6 ( 5.9 ×10 5 ) 2 = 15.7 kg / cm = + < f r 2 1.7 ×1.85 (30 )(100 ) ( 30) (100 ) 1
t ,b
The wall thicknessisoK Design for Reinforcement Long Wall
negative moment M -ve Check for small eccentricity ENGC6353
Dr. Mohammed Arafa
Page 3
e
=
M u F u
=
4.2 (100 ) 17
h
= 24.7 >
2
− d '' = 15 − 5.7 = 9.3
Small eccentricity approach can not be used h e = − d '' = 15 − 5.7 = 9.3 2 Direct tension reinforcement Ast =
T
φ f y
=
17 ×103 0.9 × 4200
= 4.5 cm 2 / m
Bending Moment Reinforcement ' M u , −ve = 4.2 − 17 × 9.3/100 = 2.6 t . m
d=30-5.7=24.3 ρ −ve
2.61⋅105 ( 2.6 ) 0.85 ⋅ 350 1 − 1 − = 0.00117 = 2 4200 100 ⋅ ( 24.3 ) ⋅ 350
As ( −ve ) = ( 0.00117 )(100 )( 24.3 ) = 2.85 cm
2
As ,total = 4.5 + 2.85 = 7.35 cm / m 2
use φ 14@20cm
Design for Positive Moment at Midspan M u' , +ve = 7.9 − 17 × 9.3/100 = 6.32 t. m
d=30-5.7=24.3 ρ +ve
2.61⋅105 ( 6.32 ) 0.85 ⋅ 350 1 − 1 − = 0.00289 = 2 4200 ⋅ ⋅ 100 24.3 350 ( )
2 As ( +ve ) = ( 0.00289 )(100 )( 24.3 ) = 7.0 cm 2 As ,total = 4.5 + 7 = 11.5 cm / m
use φ 16@15cm
Design for Short Wall
negative moment M -ve Check for small eccentricity e=
M u F u
=
5.9 (100 ) 30.2
= 19.5 >
h
2
− d '' = 15 − 5.7 = 9.3
Small eccentricity approach can not be used
ENGC6353
Dr. Mohammed Arafa
Page 4
e=
h
− d '' = 15 − 5.7 = 9.3 2 Direct tension reinforcement T
Ast =
φ f y
=
30.2 ×10
3
0.9 × 4200
= 8.0 cm 2 / m
Bending Moment Reinforcement ' M u , −ve = 5.9 − 30.2 × 9.3/100 = 3.0 t. m
d=30-5.7=24.3 ρ −ve
2.61⋅105 ( 3.0 ) 0.85 ⋅ 350 1 − 1 − = 0.00137 = 2 4200 ⋅ ⋅ 100 24.3 350 ( )
2 As ( −ve ) = ( 0.00137 )(100 )( 24.3 ) = 303 cm 2 As ,total = 8.0 + 3.3 = 11.3 cm / m
use φ 16@15cm
Design at Mid-span
Design of the Hopper Walls
The pressure changes very little with depth of the hopper, so use the pressure at the top of the hopper with Cd=1.35 1.35 × 4.65 = 6.28 t/m
2
q a ,des
=
p a ,des
=
1.35 × 0.577 × 4.65 = 3.6t/m
q b ,des
=
1.35 × 5.53 = 7.47 t/m
q b ,des
=
1.35 × 0.577 × 5.53 = 4.3t/m
2
2
2
Angleof slopes α a α a
3 = tan −1 = 48 3 − 0.3 3 = tan −1 = 60.5 2 − 0.3
2
2
2
q α a ,des
=
3.6sin 48 + 6.28cos 48 = 4.8t/m
q α b ,des
=
4.3sin 60.5 + 7.47cos 60.5 = 5.0 t/m
ENGC6353
2
2
2
Dr. Mohammed Arafa
Page 5
Horizontal Ultimate tensile forces F tau
=
1.7 ( 5.0 )( 6/2 ) sin ( 48) =19.0t/m
F tbu
=
1.7 ( 4.8 )( 4/2) sin ( 60.5) =14.2t/m
2
2
The own weight of the Hopper and its contents W L = π 3 ( 4 × 6 )( 3 )( 0.8 ) = 60 ton W L = π 3 ( 4 × 6 )( 3 )( 0.2 × 2.5 ) = 38 ton
For simplicity neglect the opening area at the bottom of the hopper. Hopper side A a and A b can be calculated as: A a = Ab =
1 2 1 2
( 4 × 3) = 6 m 2 (6 × 2) = 6 m 2
A a = Ab = 6 m
2
c a = cb = 1/ 4 F
mau =
Fmbu
=
(
1.7 c aW L
+
A aq a , des ) + 1.4c bW g a sin α a
(
1.7 cbW L
+
Ab q b ,des ) + 1.4c bW g b sin α b
=
=
1.7 ( 0.25 × 60 + 6 × 6.28 ) + 1.4 ( 0.25 )( 38 ) 4 sin 48 1.7 ( 0.25 × 60 + 6 × 7.47 ) + 1.4 ( 0.25 )(38 ) 6 sin 60.5
=
=
34.6 ton
22 ton
Hopper wall bending can be computed using Tables for triangular slabs: For Hopper wall A a = 4.3m c=
2
( 6.3 / 2 ) + 32 = 4.35m
a / c ≅ 1.0
From table 16.4 in Appendix At the centre of the top edge n x = -0.209 and n y =-1.255
( = 1.7 (1.255 ( 4.8 ) 4.3
) / 64 ) = 2.89 t . m
2 M xau = 1.7 0.209 ( 4.8 ) 4.3 / 64 = 0.493 t . m
M yau
2
This slab is to be designed for bending and tensile force similarly as shown above.
ENGC6353
Dr. Mohammed Arafa
Page 6
Design of the edge beam
Dowels are provided to transfer the vertical loads from hopper edge beam into the vertical walls T = Fmau sin α a = 34.6 sin 48 = 25.7 ton / m A st ,dowels =
25.7 ×103 0.9 × 4200
= 6.8 cm 2 / m
Since the edge beam is to be joining the vertical wall using dowels. The upper wall shear and horizontal components of the hopper are assumed to be in equilibrium. Thus no horizontal load is carried by the edge beam. Its only purpose is to simplif y construction. Minimum longitudinal steel and shear stirrups are provided
Vertical Wall
The vertical walls are analyzed as deep girder (strut and tie analysis can be used) to carry vertical the following vertical loads: From dowel
25.7 ton/m
Friction 1.7(19.35) = 32.9 ton/m Wall weight, 1.4(2.5)(0.3)(30)= Total =
ENGC6353
31.5 ton/m 90 ton/m
Dr. Mohammed Arafa
Page 7
