Usman Altaf

This is tricky. The sheet would be punctured only if a sharp object tears through it. Surface area wind might not be able to puncture plastic. It might cause some permanent deformation (strain based) but the thing might stick there.
Enclosed or partially enclosed sheds also have internal pressure that adds to members in addition to externally applied pressure. 
There is no horizontal and vertical bracing provided for the structure . The structure might be unstable. There should be some bracing no matter how small or big the wind load is.
You can get these values from the manufacturer. If the manufacturer that you are using has no data like that you can google to see what other manufacturers provide an establish a basis for your structure design.
Out of curiosity, how are these poles connected with the structure below? How are you resisting the tension due to wind.
 
 
 

Hi Rifat,
You can refer Section 4-10 of Book 'Foundation Analysis and Design, Joseph E. Bowles' for 
a complete discussion on Bearing Capacity based on SPT values.
Thanks

Hi Rifat,
You can refer Section 4-10 of Book 'Foundation Analysis and Design, Joseph E. Bowles' for 
a complete discussion on Bearing Capacity based on SPT values.
Thanks

Hi Rifat,
You can refer Section 4-10 of Book 'Foundation Analysis and Design, Joseph E. Bowles' for 
a complete discussion on Bearing Capacity based on SPT values.
Thanks

Material behavior can be idealized as consisting of an 'elastic' domain and a 'plastic' domain. For almost 200 years, structural design has been
based on an elastic theory which assumes that structures display a linear response throughout their loading history, ignoring the post-yielding
stage of behavior. Current design practice for reinforced concrete structures is a curious blend of elastic analysis to compute forces and moments, plasticity theory to proportion cross-sections for the moment and axial, load, and empirical mumbo-jumbo to proportion members for shear.
 
From the book "Design of Concrete Structures with Stress Fields" by A. Muttoni,  J. Schwartz and  B.Thurliman.
 

@miqureshi77 Bearing capacity in Lahore varies from 0.5 Tsf to 1.25Tsf at around 3ft. below natural ground level, depending on location. Generally, shallow foundations work for a maximum of 3 floors.
Thanks

Following documents may be helpful in connection with structural design of single & two-cell RC box culverts:-
1. Box Culvert Design Example-MnDOT
2. RCC BOX CULVERT - METHODOLOGY AND DESIGNS INCLUDING COMPUTER METHOD
3. Design coefficients for single and two cell box culvert
4. AASHTOWare CVT1_Two-Cell RC Box Culvert Example
More documents related to the current topic, are available here:-
Structural design of two-cell rc box culvert
Regards.

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Hi Faisal,
Please see my comments on your sketch below. I would suggest to add more shear reinforcement (at least 2 layers, one looks flimsy) between existing foundation and the jacket. Obviously, it would be based on your calculations to transfer punching shear, one way shear and moment. Also, I would suggest to add shear connectors/ dowels/ nelson studs to all vertical faces of the foundation and provide continuous reinforcement as shown connecting periphery jacket beams and new pour. Also, if you are excavating around existing foundations, please give some thought to that so that the excavation doesn't undermine adjacent foundation.
Hope that helps.
Cheers!

Well explained by Uzair.
Generally, whenever you see more reinforcement on top, it indicates either the beam is doubly reinforced or it is continous. In your case, it appears the beam is singly reinforced as nominal steel is present on top.

Yes, provided you know (or able to estimate) the concrete compressive strength & reinforcing steel yield strength.
Calculate required maximum steel area per code (corresponding to Maximum Steel Ratio in tension), for the given beam size & compare it with the actual (provided) steel area.
In case actual steel area in tension zone is more than the Code Permitted Maximum steel area, and the reinforcement area provided in compression zone is at least equal to the excess amount of reinforcement area provided in tension zone (= the difference of actually provided minus Code Maximum steel area in tension), the given beam will be a doubly reinforced beam. 
Regards.