Folded-plates

Folded plates : Folded plates are assemblies of flat plates rigidly connected together along their edges in such a way so as to make the structural system capable of carrying loads without the need for additional supporting beams along mutual edges.

Types of folded plates : 1- Prismatic : if they consist of rectangular plates. 2- Pyramidal : when non-rectangular plates are used. 3- Prismoidal, triangular or trapezoidal. On the other hand, Folded plates can be classified as: 1- single. 2- Multiple. 3- Symmetrical. 4- Unsymmetrical. 5- Simple. 6- Continuous. 7- Folded plates with simple joints. 8- Folded plates with multiple joints. 9- Folded plates with opened cross sectional. 10- Folded plates with closed cross sectional.

Historical Review :  

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The first application of folded plates back to 1924, and attributed to Ehlers in Germany. In 1932, Gruber developed a rigorous analytical solution that take into consideration the compatibility of deformations and the relative ridge displacement, but for construction of "n" number of plates, Gruber's analysis requires the analysis of 7n-5 simultaneously equations. …. Big Trouble. In 1930, Vlasov developed considerably simpler procedure that still take into consideration the compatibility of deformations and the relative ridge displacement, but for "n" number of plates, Vlasov's analysis requires the analysis of 2n simultaneously equations …. Also still hard. In 1947, an approximate and extremely simplified method was developed by Winter and Pei, in which the effect of ridge displacement was completely disregarded. but for "n" number of plates, analysis requires the analysis of n-1 simultaneously equations. Attempts were then made by a number of investigators to introduce certain corrections to Winter and Pei solution, that take into account the ridge displacements. These attempts can be classified to : Analytical methods. Iterative Techniques.

Folded Plate Behaviors : Each plate is assumed to act as a beam in its own plane, this assumption is justified when the ratio of the span "length" of the plate to its height "width" is large enough. But when this ratio is small, the plate behaves as a deep beam.

Assumptions For the analysis of Folded Plates : 1- Material is homogenous, elastic, isotropic, Hook's Law is valid, thickness of plate is small when compared to plate dimensions. 2- Problem will be treated as one-dimension if plate is assumed to behave in beam action, but in two dimensions if based on the theory of elasticity. 3- Joints are assumed to be rigid enough.

Remember : The main problem of folded plates is the analysis, not the design.

When can ridge displacements be neglected ?! Ridge displacements be neglected if there is no ridge displacement such as: - closed systems. - If the folded plate system is under symmetrical loading. - In folded plate systems with interior symmetrical planes, subjected to symmetrical loads. But in other cases ridge displacements may take place, yet they are such a small order of magnitude that their effects can be neglected, as folded plate system provided with adequate number of intermediate diaphragms.

Folded Plates With Simple Joint What do we mean by "Simple Joint", and what is the other type ? When the folded plate is that with simple joint , we mean that : no more than 2 elements are connected to the joint. But when more than 2 elements are connected to the joint, it can be named as multiple joint. The width of any plate should not be larger than 0.25 its length to be considered to act as beam.

Actions of Folded plate due to loads : 1- Slab action : loads are transmitted to ridges by the bending of plates normal to their planes. 2- Beam action : Loads are transmitted through plates in their planes to diaphragms.

** Slab Action: A strip of unit length of folded plate is taken to act as one-way slab supported at ridges, the outer plates regarded as cantilevers. The ridges are assumed to behave as rigid support, producing the reactions R2, R3, …. etc as shown in figures. and the moment diagram corresponding to slab action can be drawn as in fig.

** Beam action : Applying the reactions R2, R3, …. And resolving the load at any ridge in the directions of two adjoining plates, one can get the forces p2,1, p2,3, p3,4, …. Acting in middle planes of plates (1,2), (2,3), (3,4), … respectively. These in plane loads ansmitted by the plates behaving as beams to the supporting diaphragms.

Resolution Of Ridge Loads and Sign Conventions: Loads can be easily verified from the triangles of forces. Pi,i+1 = 𝑅𝑖+1

cos 𝛽 𝑖+1 sin 𝛼 𝑖+1

− 𝑅𝑖

cos 𝛽 𝑖−1 sin 𝛼 𝑖

Ridge loads are taken as + if they act downwards and vice versa. Plate forces are taken as + when the associated bending of the plate produces tension at the end i of the plate ( i , i+1 ).

α i : an angel to be measured in a clockwise sense from the extension of the direction ( i -1 , i ) to the direction of plate ( i , i+1 ).

βi

: represent the inclination of plate ( i , i+1 ) to the horizontal, and to be measured

in an anticlockwise direction from the horizontal line.

Contact shears : Stresses developed in the plates, due to beam action, are obtained on the basis of elementary beam theory. Therefore, such stresses vary linearly across the width of each plate. |It should be observed, however, that since the plates are connected together in monolithic manner, each plate cannot deform freely. The strain at any ridge should have the same value in each of the two plates connected to that ridge. This lead to the creation of contact shearing forces at the ridges, acting in the plates in the longitudinal direction, to equalize the normal strains at the ridge in the 2 plates connected to this ridge. For the contact shear, there are 2 approaches to solve this problem, the first is called "Equations of Three Shears", and the other is "Stress distribution method".

Three Shears Equations : To illustrate the approach, For plate (2,3) :

N = T2 – T3 ………….. (i)

M = M0 2,3 + T2 z2,3 + T3 z3,2 Where: T : is the contact shearing forces.

z : is the distance of the extreme fibers from the centroid of the cross section.

2,3 =

𝑇2 − 𝑇3

3,2 =

𝑇2 − 𝑇3

𝐴2,3 𝐴2,3

+

M 0 2,3 + T 2 z 2,3 + T 3 z 3,2 𝑍2,3

−

M 0 2,3 + T 2 z 2,3 + T 3 z 3,2

………….. (ii)

𝑍3,2

For plate (3,4) 3,4 =

𝑇3 − 𝑇4 𝐴3,4

−

M 0 3,4 + T 3 z 3,4 + T 4 z 4,3 𝑍3,4

………….. (iii)

Where : Z : is the section Modulus of the cross section.

From the compatibility of longitudinal strains: 3,2 = 3,4

………….. (iv)

Substituting from eq. (ii) and (iii) into eq. (iv), we get the relation at ridge 3. 1

1

T2 (𝐴′ ) + T3 (𝐴′′ + 2,3

2,3

1 𝐴′′3,4

1

) + T4 (𝐴′′ ) = − ( 3,4

𝑀0 2,3 𝑍3,2

+

𝑀0 3,4 𝑍3,4

………….. (v)

)

Where: 1 𝐴′2,3 1 𝐴′′2,3 1 𝐴′2,3 1 𝐴′2,3

𝑧

= Z2,3 −

1

3,2

𝐴2,3

= Z3,2 − 3,2

𝐴2,3

= Z4,3 − 3,4

𝐴3,4

= Z3,4 −

𝐴3,4

𝑧

1

𝑧

1

𝑧

3,4

1

Special Case : If the elements of the folded plate system have rectangular cross section, where the thickness may vary from element to another, equations may be simplified according to : 𝑧𝑖−1,𝑖 = 𝑧𝑖,𝑖−1 =

ℎ 𝑖−1,𝑖

,

2

Z𝑖−1,𝑖 = Z𝑖,𝑖−1

𝑧 𝑖−1,𝑖

,

Z 𝑖−1,𝑖

=

3 𝐴𝑖−1,𝑖

And equation (v), will reduce to the form: Ti-1 (

1

𝐴𝑖−1,𝑖

) + 2Ti (

1

𝐴𝑖−1,𝑖

+

1 𝐴𝑖,𝑖+1

) + Ti+1 (

1

𝐴𝑖,𝑖+1

)=−

1 2

(

𝑀0 𝑖−1,𝑖 𝐙𝑖,𝑖−1

+

𝑀0 𝑖,𝑖+1 𝐙𝑖,𝑖+1

)

Then the stresses can be calculated from equations (ii), and (iii) … After calculating stresses, design is an easy procedure.

Stress Distribution Method : This method is easy and similar to the moment distribution technique, with fiber stresses in individual plates due to the in-plane moment M 0 "free edge stresses" represent the counterpart of the fixed end moment dealt with in moment distribution technique. Stress can be calculated according to : 4 𝑇𝑖

 i,i-1 = − 𝐴

𝑖−1,𝑖

,

4 𝑇𝑖

 i,i+1 = 𝐴

𝑖,𝑖+1

Stiffness factor S as the stress induced due to a unit value of T, S i,i-1 = − 𝐴

4

,

𝑖−1,𝑖

S i,i+1 = 𝐴

4 𝑖,𝑖+1

Distribution factors at ridge I, where plates (i-1,i) and (i,i+1) meet are given by: d i,i-1 = − 𝐴

𝐴𝑖,𝑖+1 𝑖−1,𝑖 + 𝐴 𝑖,𝑖+1

,

d i,i+1 = 𝐴

𝐴𝑖−1,𝑖 𝑖−1,𝑖

+ 𝐴𝑖,𝑖+1

Stresses corresponding to T, at far ends of the two adjoining plates are:  i-1,1 =

2 𝑇𝑖 𝐴𝑖−1,𝑖

,

 i+1,i = −

2 𝑇𝑖 𝐴𝑖,𝑖+1

The Carry over factors, C, will be given by: 1

C i,i-1 = C i,i+1 = − 2 But for exterior plates, C 2,1 =

1 2

, and C 1,2 = 0.

Example: All elements of the folded plate roof shown in the figure have a constant thickness of 10cm. Taking the unit weight of concrete to be 2.5 t/m3, and assuming the roof to be subjected to a superimposed vertical load of intensity 100 Ksm of horizontal projection, find the distribution of longitudinal stresses at the mid-span section. i) ii)

Analytically, by writing down the equations of three shears. Numerically, using the method of stress distribution.

Solution:. Due to symmetry, one has to analyze only half of the cross section of the roof. Plate 1,2 2,3 3,4

h “m” 0.750 3.000 3.304

t “m” 0.1 0.1 0.1

A “ m2 ” 0.0750 0.3000 0.3304

Z “ m3 ” 0.009375 0.150000 0.181940