In
Figure A1, we can see the loading condition of the complete S-structure and of one Representative Unit Cell Element (RUCE). In this analytical study, parts of the S-structure are assumed to be Timoshenko beams (TB). Various parts of the S-shaped portion of a RUCE are shown in
Figure A2 and represented by elastic lines. These are the parts of the RUCE that are considered TB. There are two struts at the left and right that connect S-shaped parts, as can be seen in
Figure A1(a) and
Figure A1(b). Parts 1 and 3 deform in a similar way, so both the parts are represented by one free-body diagram (FBD), and part 2 deforms differently, so one FBD represents this part, as can be seen from
Figure A3 and
Figure A4. It may seem as if there is an unbalanced moment, but this is balanced due to its connection to adjacent RUCEs, which have an unbalanced moment of opposite signs, and they contribute equally to balance the unbalanced moment generated in a RUCE.
Figure A1.
(a)Loading condition of S-structure and (b) one RUCE of S-structure under transverse loading
Figure A2.
Various parts of the S-shaped portion of the RUCE
Figure A3.
(a) FBD of parts 1 and 3 and (b) shear force and (c) bending moment on a section of it
Appendix A.1.1. Shear Force and Bending MOMENT distribution
Parts 1 and 3:
For parts 1 and 3, the free-body diagram is represented by the leftmost figure of
Figure A3. The shear force at various locations of the part is represented by the middle one, and the bending moment at various locations of the parts is represented in the rightmost figure. At the fixed end of parts 1 and 3, the supporting moment is
, and the supporting load is
. At the junction of struts (that connect two RUCEs) and arc-like parts, a balancing moment of
is applied.
Shear force distribution:
Let’s say the line along part 2 represents , and is positive in the clockwise direction. Now, for , shear force, ; and for , shear force, . For , , it is assumed that the horizontal displacement of all the RUCEs in a column is the same.
Bending moment distribution:
For , bending moment, =-, and for , bending moment, =[].For , =0.
Part 2:
The sub-figures in
Figure A4 are arranged for part 2 of the RUCE like those in
Figure A3 for parts 1 and 3. At the junction of the inclined part and the arc-like parts, the support moment,
, and the supporting load is
.
Shear force distribution:
Shear force, at any x=x, , where `x’ is the distance along the length of part 2.
Bending moment distribution:
For x=x, bending moment, , where .
Figure A4.
(a) FBD of part 2 and (b) bending moment and (c) shear force on a section of it
Figure A4.
(a) FBD of part 2 and (b) bending moment and (c) shear force on a section of it
Appendix A.1.2. Analytical Investigation of Elastic Modulus
The total vertical deflection is given as . Again, the deflection of any part is the summation of deflection due to shearing. Due to bending, e.g., , where is the vertical deflection due to shear and is due to bending.
Parts 1 and 3:
Shear stress distribution on a section at any angle
is obtained as,
and after simplifying it, it is obtained that,
Again, shear stress distribution on a section at any angle
is obtained as,
and again, after simplifying it, it is obtained that,
Now, strain energy due to shear for part 1,
So, using Castigliano’s second theorem,
Again, strain energy due to bending for part 1 is given by,
So, using Castigliano’s second theorem,
Part 2:
Shear stress distribution on part 2 at any given x=x is given by,
Strain energy due to shear on part 2 is,
So, using Castigliano’s second theorem,
On the other hand, the strain energy due to bending in part 2 is,
So, using Castigliano’s second theorem,
left and right struts:
The left and right struts deform axially in a similar manner, so the axial deflection of the left strut is,
Total axial deflection of the left and right struts, , where is the axial deflection of the right strut, is the axial deflection of the left strut, and .
Total Deflection of RUCE:
Total horizontal deflection of a RUCE in
row and
column in the structure is,
RUCE to Entire System:
Figure A5.
Assembly of RUCEs to obtain entire system
Figure A5.
Assembly of RUCEs to obtain entire system
The method of assembly is given in
Figure A5. First, RUCEs in one column are combined, and then multiple columns are assembled to get the entire system.
Now, it is known that the force on all the RUCEs in a single column is the same, or along the column same force is experienced by all RUCEs, i.e.,
. So,
as there are m RUCEs in a column. For simplification, it is assumed that,
It is known that all the columns deform equally, and this is equal to the total deflection of the entire structure, i.e.,
. To achieve static equilibrium, the following must be true-
Again,
and
. So, effective transverse elastic modulus,
. It should be noted that all cross sections are rectangular with
. Putting all of these in the equation, it is obtained that,
Vertical Deflection of RUCE:
Figure A6.
Deformed (a)part 1 or 3 and (b) part 2 of RUCE: continuous line represents undeformed situation and dashed line represents deformed situation
Figure A6.
Deformed (a)part 1 or 3 and (b) part 2 of RUCE: continuous line represents undeformed situation and dashed line represents deformed situation
The total horizontal deflection of a RUCE,
and the total vertical deflection of a RUCE,
From here, the value of
is obtained as,
Similarly,
. Now, from
Figure A6(b), it can be said that,
From here,
can be evaluated.
So, from equation
A25, it is obtained that,
A.1.3. Poisson’s Ratio Calculation:
and
where x is along the horizontal direction, and y is along the vertical direction, as mentioned in
Figure 2. Finally, Poisson’s Ratio for one RUCE is obtained as,
(notice), ; where, and .
Now, the vertical deflection of the mid-plane of the S-structure is considered in the expression of the Poisson’s ratio. Hence, for a column,
where
is strain in x direction averaged over
RUCEs; or,
As we know,
, and we assume that
, where
is the average strain of the unit cells in the `
’ row in the y-direction. upon putting these values in the expression, it is obtained that,
Further, it is assumed that
and the total contraction of the mid-plane is obtained as,
Or,
where
,
and
are Poisson’s ratio, strain along x direction and strain along y-direction of the mid-plane of the overall structure. It is known that all the columns deflect by the same distance or all columns have the same strain in the x-direction, i.e.,
Or,
as
is the same for every column. Now,
Where is dependent on .