CIRCULAR PLATES The deflection profile of a circular plate subjected to uniformly distributed load covering the entire surface (symmetric about an axis perpendicular to the plate through its centre) will be symmetrical about its centre O. The deflections will be the same at points equidistant from the plate centre. Thus, it is adequate to to consider the deflections along a diameter of the plate domain as shown in Figure 4.
B
φ
O w
A dr
r Z
Figure 4 Bending of a circular plate about a diameter
Consider a circular plate with origin at its centre; r denotes the radial distance of a point, φ is the angle subtended by normal to the deflected surface with OB, the axis of symmetry, while w denotes the vertical deflection (in the Z – direction) of the middle plane of the plate. The rotation (slope ) of any point due to the applied load is given by φ = - ∂w / ∂r, while the curvature for small deflections at is given by 2
1 / r n = - ∂ w / ∂r 2 =
∂φ
/ ∂r
[ 1]
Because of the symmetry, 1 / r n is one of the principal curvatures of the deflection surface at A. The second principal curvature will be in the section through normal AB, and perpendicular to the r Z plane. 1
The normals such as AB, for all points of the middle surface at radial distance r , form a conical surface with apex B. Thus AB is the radius of the second principal curvature r t . From the figure, we have 1 / r t = - ( 1 / r ) ( ∂w / ∂r ) = φ / r [ 2] The bending moments at A can be written as Mr = - D [ ( = D[(
2
2
w / ∂r ) + υ ( 1 / r ) ( ∂w / ∂r ) ] ∂ φ / ∂r ) + υ ( φ / r ) ]
∂
[ 3]
2
Mt = - D [ ( 1 / r ) ( ∂w / ∂r ) + υ ( ∂ w / ∂r 2)] = D [ ( φ / r ) + υ ( ∂ φ / ∂r )]
[ 4]
Q + ( dQ / dr ) dr Mr + ( dMr / dr) dr
Mr O
Q dr
b
r
Z
c dθ
r
O d a Figure 5 Forces on an element of a circular plate
Mr acts along the circumferential sections of the plate in the r - Z plane, while Mt acts along the diametral section r - Z of the plate. The equilibrium of an element a b c d of the plate is considered; the moments and shear force on the elements are noted in the figure shown below. The element subtends an angle dθ at the centre of the plate. The moment on the length cd of the element is given by Mr r dθ; the corresponding moment on the side ab is given by [ Mr + ( dMr / dr) dr ] [ ( r + dr) dθ ]
2
Similarly, the moments on the sides ab and cd of the element are Mt dr; their component in the plane r – O – Z is given by Mt dr dθ. Because of the symmetry, the shear forces on the diametral sections of the plate (sides bc and ad) vanish, while those on the circumferential planes will balance the applied loads. If Q is the shear force per unit length of the cylindrical section of radius r (side cd of the element), the total shear force on side cd is given by Q r dθ dr. The corresponding shear force on the side ab of the element is given by [ Q + ( dQ / dr ) dr ] ( r + dr ) dθ. These forces induce a moment [ Q r dr dθ ] on the element. The equilibrium condition for the moments is given by [ Mr + ( dMr / dr) dr ] [ ( r + dr) dθ ] - Mr dr dθ - Mt dr dθ + [ Q r dr dθ ] = 0 Simplifying the above equation, we obtain Mr + ( dMr / dr) r - Mt + Q r = 0 Substituting for Mr and Mt from Eqn. 3 and 4, we obtain
or
2
(
∂
(
∂
3
φ / ∂r 2 ) + ( 1 / r ) ( 3
w / ∂r ) + ( 1 / r ) (
∂ ∂
φ / ∂r ) - ( φ / r 2) = - Q / D 2
2
w / ∂r ) - (
∂
φ / ∂r ) ( 1 / r 2) = Q / D
[ 5]
These equations can be written in the form, ( d / dr ) [( 1 / r ) (d / d r ) ( r φ ) ] = - Q / D ( d / dr ) [( 1 / r ) (d / dr ) (r d w/ dr ) ] = Q / D
[ 6]
The value of Q can be computed by considering the load within a circle of radius r, and dividing the load by the circumference (2 π r). The deflection profile of the circular plate for a given loading can be obtained by integrating Eqn. 6.
Circular plate under udl
Consider a circular plate of radius a, subjected to a udl q over the entire surface. The shear force Q at a radial distance r from the centroid of the circle is given by 2 π r Q = π r 2 q
or Q = q r / 2
Substituting in the equilibrium equation for the plate (Equation 6), we obtain ( d / dr ) [( 1 / r ) ( d / dr ) (r d w/ dr ) ] = Q / D = (qr/2)/D
3
Integrating the equation successively, we obtain [( 1 / r ) ( d / d r ) ( r d w/ dr ) ] = q r 2 / ( 4 D ) + C1 Multiplying both sides of the equation by r, and integrating again, we obtain 2
[ ( r d w/ dr ) ] = q r 4 / ( 16 D ) + ( C 1 r / 2 ) + C2 Simplifying the above equation, d w/ dr = q r 3 / ( 16 D ) + ( C1 r / 2 ) + C2 / r Integrating again, we obtain 2
w = q r 4 / ( 64 D ) + ( C 1 r / 4 ) + C2 ln ( r / a) + C3
[ 7]
The constants of integration can be determined by substituting the boundary conditions. Circular plate with clamped edge
The boundary conditions in this case are
and
w = 0 and dw / dr = 0 dw / dr = 0
at r = a at r = 0
because of symmetry
dw / dr = q r 3 / ( 16 D ) + ( C1 r / 2 ) + C2 / r From the boundary condition on the slope at the plate centre (r = 0), we have C2 = 0; the boundary condition at r = a yields
q a 3 / ( 16 D ) + ( C1 a / 2 ) = 0
or
C1 = - q a 2 / ( 8 D )
Substituting for the constants in Eqn. 7, we obtain 2
w = q r 4 / ( 64 D ) - [ q a 2 r / ( 32 D ) ] + C3
From the boundary condition w = 0 at r = 0, we have
or
w = q a 4 / ( 64 D ) - [ q a 4 / ( 32 D ) ] + C3 = 0 C3 = + q a 4 / ( 64 D )
The deflection profile of the plate is given by 2
w = [ q r 4 / ( 64 D ) ] - [ q a 2 r / ( 32 D ) ] + [ q a 4 / ( 64 D ) ] = [ q (a 2 - r 2 )2 / ( 64 D ) ]
4
[ 8]
The maximum deflection of the plate at the centre ( r = 0) is given by wmax = [ q a 4 / ( 64 D ) ]
[ 9]
The bending moments can be obtained by substituting for w in Eqn. 3 and 4. We have w = [ q (a 2 - r 2 )2 / ( 64 D ) ] ∂w 2
∂
/ ∂r = - [ q r (a 2 - r 2 ) / ( 16 D ) ] 2
w / ∂r = - [ q ( a 2 - 3 r 2 ) / ( 16 D ) ] 2
2
Mr = - D [ ( ∂ w / ∂r ) + υ ( 1 / r ) ( ∂w / ∂r ) ] = + D [ { q ( a 2 - 3 r 2 ) / ( 16 D ) } + ( υ / r ) { q r (a 2 - r 2 ) / ( 16 D ) }] 2 2 = ( q / 16) [ a ( 1 + υ) - r ( 3 + υ) ] 2
Mt = - D [ ( 1 / r ) ( ∂w / ∂r ) + υ ( ∂ w / ∂r 2)] = + D [( 1 / r ) { q r (a 2 - r 2 ) / ( 16 D ) } + υ { q ( a 2 - 3 r 2 ) / ( 16 D )}] 2 2 = ( q / 16) [ a ( 1 + υ) - r ( 1 + 3 υ) ]
[ 10 ]
The bending moments at the centre of the plate can be computed by substituting r = 0; we have 2
Mr = ( q / 16) [ a ( 1 + υ)] = Mt
[ 11 ]
The bending moments at the edge of the plate can be computed by substituting r = a ; we have 2
Mr = - q a / 8
and
2 Mt = - q a υ / 8
5
[ 12 ]
PROBLEM I A roof slab of a reinforced concrete cylindrical reservoir is 2.0 m in radius and is subjected to a udl of 1.5 kPa (including the self weight of the slab). The slab is cast monolithic with the reservoir wall, and is assumed as rigidly fixed. Derive expressions for the radial and circumferential moments in the slab. If the stress in the slab is not to exceed 1.4 MPa, calculate i ii
minimum thickness of the slab 2 maximum deflection of the slab, if its flexural rigidity is 100.0 kN m / m.
Consider a circular plate of radius a, subjected to a udl q over the entire surface. The shear force Q at a radial distance r from the centroid of the circle is given by 2 π r Q = π r 2 q
or Q = q r / 2
Substituting in the equilibrium equation for the plate (Equation 6), we obtain ( d / dr ) [( 1 / r ) ( d / dr ) (r d w/ dr ) ] = Q / D = (qr/2)/D Integrating the equation successively, we obtain [( 1 / r ) ( d / d r ) ( r d w/ dr ) ] = q r 2 / ( 4 D ) + C1 Multiplying both sides of the equation by r, and integrating again, we obtain [ ( r d w/ dr ) ] = q r 4 / ( 16 D ) + ( C 1 r 2 / 2 ) + C2 Simplifying the above equation, d w/ dr = q r 3 / ( 16 D ) + ( C1 r / 2 ) + C2 / r Integrating again, we obtain 2
w = q r 4 / ( 64 D ) + ( C 1 r / 4 ) + C2 ln ( r / a) + C3
The constants of integration can be determined by substituting the boundary conditions. Circular plate with clamped edge
The boundary conditions in this case are w = 0 and dw / dr = 0
at r = a 6
and
dw / dr = 0
at r = 0
because of symmetry
dw / dr = q r 3 / ( 16 D ) + ( C1 r / 2 ) + C2 / r From the boundary condition on the slope at the plate centre (r = 0), we have C2 = 0; the boundary condition at r = a yields
q a 3 / ( 16 D ) + ( C1 a / 2 ) = 0
C1 = - q a 2 / ( 8 D )
or
Substituting for the constants in Eqn. 7, we obtain 2
w = q r 4 / ( 64 D ) - [ q a 2 r / ( 32 D ) ] + C3
From the boundary condition w = 0 at r = 0, we have
or
w = q a 4 / ( 64 D ) - [ q a 4 / ( 32 D ) ] + C3 = 0 C3 = + q a 4 / ( 64 D )
The deflection profile of the plate is given by 2
w = [ q r 4 / ( 64 D ) ] - [ q a 2 r / ( 32 D ) ] + [ q a 4 / ( 64 D ) ] = [ q (a 2 - r 2 )2 / ( 64 D ) ]
The maximum deflection of the plate at the centre ( r = 0) is given by wmax = [ q a 4 / ( 64 D ) ] The bending moments can be obtained by substituting for w in Eqn. 3 and 4. We have w = [ q (a 2 - r 2 )2 / ( 64 D ) ] ∂w 2
∂
/ ∂r = - [ q r (a 2 - r 2 ) / ( 16 D ) ] 2
w / ∂r = - [ q ( a 2 - 3 r 2 ) / ( 16 D ) ] 2
2
Mr = - D [ ( ∂ w / ∂r ) + υ ( 1 / r ) ( ∂w / ∂r ) ] = + D [ { q ( a 2 - 3 r 2 ) / ( 16 D ) } + ( υ / r ) { q r (a 2 - r 2 ) / ( 16 D ) }] 2 2 = ( q / 16) [ a ( 1 + υ) - r ( 3 + υ) ] 2
2
Mt = - D [ ( 1 / r ) ( ∂w / ∂r ) + υ ( ∂ w / ∂r )] = + D [( 1 / r ) { q r (a 2 - r 2 ) / ( 16 D ) } + υ { q ( a 2 - 3 r 2 ) / ( 16 D )}] 2 2 = ( q / 16) [ a ( 1 + υ) - r ( 1 + 3 υ) ]
7
The bending moments at the centre of the plate can be computed by substituting r = 0; we have 2
Mr = ( q / 16) [ a ( 1 + υ)] = Mt The bending moments at the edge of the plate can be computed by substituting r = a ; we have Mr = - q a2 / 8
and
For the given slab q = 1.5 kPa,
2
Mt = - q a υ / 8
a = 2.0 m,
σ = 1.4 MPa and
The design bending moment for the slab (at the edge) 2
Mr = - q a / 8 2 = - 1.5 X 2.0 / 8 = - 0.75 kN m / m 3 = - 0.75 X 10 N mm / mm The stress in a rectangular section width b and thickness h is given by
σ= M/Z
2
= 6 M / (b h ) or
h = √ [ 6 M / ( σ b )] 3 = √ [ (6 X 0.75 X 10 ) / (1.4 X 1.0) ] = 56.7 mm
The minimum thickness of the slab required = 56.7 mm The maximum deflection in the slab is given by wmax = = = =
[ q a 4 / ( 64 D ) ] 4 [ 1.5 X 2.0 / ( 64 X 100.0) ] -3 3.75 X 10 m 3.75 mm
8
2
D = 100.0 kN m / m
PROBLEM II Compute the maximum deflection in a circular slab of 15.0 m diameter and 0.20 m thickness, when subjected to a uniformly distributed load of 10.0 kPa. Assume fixed edge conditions, and adopt Young’s modulus of 25.0 GPa and Poisson’s ratio of 0.20.
For the given slab, we have a = 7.5 m, q = 10.0 kPa, h = 0.20 m, E = 25.0 GPa, υ = 0.20 3
2
D = E h / [ 12 (1 - υ ] 6 3 2 3 2 = 25.0 X 10 X 0.2 / [ 12 (1 – 0.2 ) ] = 17.36 X 10 kN m / m 4
wmax = [ q a / ( 64 D ) ] 4 3 = 10.0 X ( 7.5) / (64 X 17.36 X 10 ) -3 = 28.48 X 10 m = 28.48 mm
9
