S ECTION A3 T HIN P LATE T HEORY
7.6 Energy method for the bending of thin plates
7.6.3 Potential energy of in-plane loads
We may consider each load Nx, Ny and Nxyin turn, then use the principle of super- position to determine the potential energy of the loading system when they act simultaneously. Consider an elemental strip of width δy along the length a of the plate in Fig. 7.15(a). The compressive load on this strip is Nxδy and due to the bending of the plate the horizontal length of the strip decreases by an amountλ, as shown in
7.6 Energy method for the bending of thin plates 243
Fig. 7.15(a) In-plane loads on plate; (b) shortening of element due to bending.
Fig. 7.15(b). The potential energy δVx of the load Nxδy, referred to the undeflected position of the plate as the datum, is then
δVx = −Nxλδy (7.40)
From Fig. 7.15(b) the length of a small elementδa of the strip is δa=(δx2+δw2)12
and since∂w/∂x is small then
δa≈δx
1+ 1 2
∂w
∂x 2
Hence
a= a
0
1+1
2 ∂w
∂x 2
dx
giving
a=a+ a
0
1 2
∂w
∂x 2
dx and
λ=a−a= a
0
1 2
∂w
∂x 2
dx Since
a
0
1 2
∂w
∂x 2
dx only differs from a
0
1 2
∂w
∂x 2
dx by a term of negligible order we write
λ= a
0
1 2
∂w
∂x 2
dx (7.41)
The potential energy Vxof the Nxloading follows from Eqs (7.40) and (7.41), thus Vx = −1
2 a
0
b
0
Nx ∂w
∂x 2
dx dy (7.42)
Similarly
Vy= −1 2
a
0
b
0
Ny ∂w
∂y 2
dx dy (7.43)
The potential energy of the in-plane shear load Nxymay be found by considering the work done by Nxyduring the shear distortion corresponding to the deflection w of an element. This shear strain is the reduction in the right angle C2AB1to the angle C1AB1 of the element in Fig. 7.16 or, rotating C2A with respect to AB1to AD in the plane C1AB1, the angle DAC1. The displacement C2D is equal to (∂w/∂y)δy and the angle DC2C1is∂w/∂x. Thus C1D is equal to
∂w
∂x
∂w
∂yδy
and the angle DAC1 representing the shear strain corresponding to the bending displacement w is
∂w
∂x
∂w
∂y
so that the work done on the element by the shear force Nxyδx is 1
2Nxyδx∂w
∂x
∂w
∂y
7.6 Energy method for the bending of thin plates 245
Fig. 7.16Calculation of shear strain corresponding to bending deflection.
Similarly, the work done by the shear force Nxyδy is 1
2Nxyδy∂w
∂x
∂w
∂y and the total work done taken over the complete plate is
1 2
a
0
b
0
2Nxy∂w
∂x
∂w
∂ydx dy
It follows immediately that the potential energy of the Nxyloads is Vxy= −1
2 a
0
b
0
2Nxy∂w
∂x
∂w
∂ydx dy (7.44)
and for the complete in-plane loading system we have, from Eqs (7.42), (7.43) and (7.44), a potential energy of
V = −1 2
a
0
b
0
Nx
∂w
∂x 2
+Ny ∂w
∂y 2
+2Nxy∂w
∂x
∂w
∂y
dx dy (7.45) We are now in a position to solve a wide range of thin plate problems provided that the deflections are small, obtaining exact solutions if the deflected form is known or approximate solutions if the deflected shape has to be ‘guessed’.
Considering the rectangular plate of Section 7.3, simply supported along all four edges and subjected to a uniformly distributed transverse load of intensity q0, we know that its deflected shape is given by Eq. (7.27), namely
w=∞
m=1
∞ n=1
Amnsinmπx
a sinnπy b
The total potential energy of the plate is, from Eqs (7.37) and (7.39) U+V =
a
0
b
0
D 2
∂2w
∂x2 +∂2w
∂y2 2
−2(1−ν)
∂2w
∂x2
∂2w
∂y2 − ∂2w
∂x∂y 2
−wq0
dx dy (7.46)
Substituting in Eq. (7.46) for w and realizing that ‘cross-product’ terms integrate to zero, we have
U+V = a
0
b
0
D
2 ∞ m=1
∞ n=1
A2mn
π4 m2
a2 +n2 b2
2
sin2mπx
a sin2nπy b
−2(1−ν)m2n2π4 a2b2
sin2mπx
a sin2nπy
b −cos2mπx
a cos2nπy b
−q0 ∞ m=1
∞ n=1
Amnsinmπx
a sinnπy b
dx dy
The term multiplied by 2(1−ν) integrates to zero and the mean value of sin2or cos2 over a complete number of half waves is 12, thus integration of the above expression yields
U+V= D 2
∞ m=1,3,5
∞ n=1,3,5
A2mnπ4ab 4
m2 a2 + n2
b2 2
−q0 ∞ m=1,3,5
∞ n=1,3,5
Amn 4ab π2mn
(7.47) From the principle of the stationary value of the total potential energy we have
∂(U+V )
∂Amn = D
22Amnπ4ab 4
m2 a2 + n2
b2 2
−q0 4ab π2mn =0 so that
Amn= 16q0
π6Dmn[(m2/a2)+(n2/b2)]2 giving a deflected form
w= 16q0 π6D
∞ m=1,3,5
∞ n=1,3,5
sin (mπx/a) sin (nπy/b) mn[(m2/a2)+(n2/b2)]2 which is the result obtained in Eq. (i) of Example 7.1.
The above solution is exact since we know the true deflected shape of the plate in the form of an infinite series for w. Frequently, the appropriate infinite series is not known so that only an approximate solution may be obtained. The method of solution, known
7.6 Energy method for the bending of thin plates 247 as the Rayleigh–Ritz method, involves the selection of a series for w containing a finite number of functions of x and y. These functions are chosen to satisfy the boundary conditions of the problem as far as possible and also to give the type of deflection pattern expected. Naturally, the more representative the ‘guessed’ functions are the more accurate the solution becomes.
Suppose that the ‘guessed’ series for w in a particular problem contains three different functions of x and y. Thus
w=A1f1(x, y)+A2f2(x, y)+A3f3(x, y)
where A1, A2and A3are unknown coefficients. We now substitute for w in the appropri- ate expression for the total potential energy of the system and assign stationary values with respect to A1, A2and A3in turn. Thus
∂(U+V )
∂A1 =0 ∂(U+V )
∂A2 =0 ∂(U+V )
∂A3 =0 giving three equations which are solved for A1, A2and A3.
Example 7.4
A rectangular plate a×b, is simply supported along each edge and carries a uniformly distributed load of intensity q0. Assuming a deflected shape given by
w=A11sinπx a sinπy
b
determine the value of the coefficient A11 and hence find the maximum value of deflection.
The expression satisfies the boundary conditions of zero deflection and zero curvature (i.e. zero bending moment) along each edge of the plate. Substituting for w in Eq. (7.46) we have
U+V= a
0
b
0
DA211 2
π4
(a2b2)2(a2+b2)2sin2 πx
a sin2πy
b −2(1−ν)
× π4
a2b2sin2πx
a sin2πy b − π4
a2b2cos2πx
a cos2 πy b
−q0A11sinπx a sinπy
b
dx dy whence
U+V= DA211 2
π4
4a3b3(a2+b2)2−q0A114ab π2 so that
∂(U+V )
∂A11 = DA11π4
4a3b3 (a2+b2)2−q04ab π2 =0
and
A11= 16q0a4b4 π6D(a2+b2)2 giving
w= 16q0a4b4
π6D(a2+b2)2sinπx a sinπy
b At the centre of the plate w is a maximum and
wmax= 16q0a4b4 π6D(a2+b2)2 For a square plate and assumingν=0.3
wmax=0.0455q0 a4 Et3 which compares favourably with the result of Example 7.1.
In this chapter we have dealt exclusively with small deflections of thin plates. For a plate subjected to large deflections the middle plane will be stretched due to bending so that Eq. (7.33) requires modification. The relevant theory is outside the scope of this book but may be found in a variety of references.
References
1 Jaeger, J. C., Elementary Theory of Elastic Plates, Pergamon Press, New York, 1964.
2 Timoshenko, S. P. and Woinowsky-Krieger, S., Theory of Plates and Shells, 2nd edition, McGraw- Hill Book Company, New York, 1959.
3 Timoshenko, S. P. and Gere, J. M., Theory of Elastic Stability, 2nd edition, McGraw-Hill Book Company, New York, 1961.
4 Wang, Chi-Teh, Applied Elasticity, McGraw-Hill Book Company, New York, 1953.
Problems
P.7.1 A plate 10 mm thick is subjected to bending moments Mxequal to 10 Nm/mm and Myequal to 5 Nm/mm. Calculate the maximum direct stresses in the plate.
Ans. σx,max= ±600 N/mm2, σy,max= ±300 N/mm2.
P.7.2 For the plate and loading of problem P.7.1 find the maximum twisting moment per unit length in the plate and the direction of the planes on which this occurs.
Ans. 2.5 N m/mm at 45◦to the x and y axes.
P.7.3 The plate of the previous two problems is subjected to a twisting moment of 5 Nm/mm along each edge, in addition to the bending moments of Mx=10 N m/mm
Problems 249 and My=5 N m/mm. Determine the principal moments in the plate, the planes on which they act and the corresponding principal stresses.
Ans. 13.1 N m/mm, 1.9 N m/mm, α= −31.7◦, α= +58.3◦, ±786 N/mm2,
±114 N/mm2.
P.7.4 A thin rectangular plate of length a and width 2a is simply supported along the edges x=0, x=a, y= −a and y= +a. The plate has a flexural rigidity D, a Poisson’s ratio of 0.3 and carries a load distribution given by q(x, y)=q0sin(πx/a). If the deflection of the plate may be represented by the expression
w= qa4 Dπ4
1+A coshπy a +Bπy
a sinhπy a
sinπx
a determine the values of the constants A and B.
Ans. A= −0.2213, B=0.0431.
P.7.5 A thin, elastic square plate of side a is simply supported on all four sides and supports a uniformly distributed load q. If the origin of axes coincides with the centre of the plate show that the deflection of the plate can be represented by the expression
w= q
96(1−ν)D[2(x4+y4)−3a2(1−ν)(x2+y2)−12νx2y2+A]
where D is the flexural rigidity,νis Poisson’s ratio and A is a constant. Calculate the value of A and hence the central deflection of the plate.
Ans. A=a4(5−3ν)/4, Cen. def.=qa4(5−3ν)/384D(1−ν)
P.7.6 The deflection of a square plate of side a which supports a lateral load represented by the function q(x, y) is given by
w(x, y)=w0cosπx
a cos3πy a
where x and y are referred to axes whose origin coincides with the centre of the plate and w0is the deflection at the centre.
If the flexural rigidity of the plate is D and Poisson’s ratio isνdetermine the loading function q, the support conditions of the plate, the reactions at the plate corners and the bending moments at the centre of the plate.
Ans. q(x, y)=w0D100π4 a4 cosπx
a cos3πy a The plate is simply supported on all edges.
Reactions:−6w0D π
a 2
(1−ν) Mx=w0D
π a
2
(1+9ν), My=w0D π
a 2
(9+ν).
P.7.7 A simply supported square plate a×a carries a distributed load according to the formula
q(x, y)=q0x a
where q0is its intensity at the edge x=a. Determine the deflected shape of the plate.
Ans. w= 8q0a4 π6D
∞ m=1,2,3
∞ n=1,3,5
(−1)m+1
mn(m2+n2)2sinmπx
a sinnπy a
P.7.8 An elliptic plate of major and minor axes 2a and 2b and of small thickness t is clamped along its boundary and is subjected to a uniform pressure difference p between the two faces. Show that the usual differential equation for normal displacements of a thin flat plate subject to lateral loading is satisfied by the solution
w=w0
1− x2 a2 − y2
b2 2
where w0is the deflection at the centre which is taken as the origin.
Determine w0in terms of p and the relevant material properties of the plate and hence expressions for the greatest stresses due to bending at the centre and at the ends of the minor axis.
Ans. w0= 3p(1−ν2) 2Et3
3 a4 + 2
a2b2 + 3 b4
Centre, σx,max= ±3pa2b2(b2+νa2)
t2(3b4+2a2b2+3a4), σy,max= ±3pa2b2(a2+νb2) t2(3b4+2a2b2+3a4) Ends of minor axis
σx,max= ±6pa4b2
t2(3b4+2a2b2+3a4), σy,max= ±6pb4a2 t2(3b4+2a2b2+3a4)
P.7.9 Use the energy method to determine the deflected shape of a rectangular plate a×b, simply supported along each edge and carrying a concentrated load W at a position (ξ,η) referred to axes through a corner of the plate. The deflected shape of the plate can be represented by the series
w=∞
m=1
∞ n=1
Amnsinmπx
a sinnπy b
Ans. Amn= 4W sinmπξ
a sinnπη b π4Dab[(m2/a2)+(n2/b2)]2
Problems 251 P.7.10 If, in addition to the point load W , the plate of problem P.7.9 supports an in-plane compressive load of Nxper unit length on the edges x=0 and x=a, calculate the resulting deflected shape.
Ans. Amn= 4W sinmπξ
a sinnπη b abDπ4
m2 a2 +n2
b2 2
− m2Nx π2a2D
P.7.11 A square plate of side a is simply supported along all four sides and is subjected to a transverse uniformly distributed load of intensity q0. It is proposed to determine the deflected shape of the plate by the Rayleigh–Ritz method employing a
‘guessed’ form for the deflection of w=A11
1− 4x2
a2 1−4y2 a2
in which the origin is taken at the centre of the plate.
Comment on the degree to which the boundary conditions are satisfied and find the central deflection assumingν=0.3.
Ans. 0.0389q0a4 Et3
P.7.12 A rectangular plate a×b, simply supported along each edge, possesses a small initial curvature in its unloaded state given by
w0=A11sinπx a sinπy
b
Determine, using the energy method, its final deflected shape when it is subjected to a compressive load Nx per unit length along the edges x=0, x=a.
Ans. w= A11
1−Nxa2
π2D 1+a2 b2
2sinπx a sinπy
b