Not a simple question as it depends on the shape of the object (drag coefficient) and the density of air as well as speed.
F = 0.5 * rho * v^2 * Cd * A
where rho is the density of air, and has to be calculated as a function of absolute barometric pressure (not corrected to sea level) temperature, and humidity (humid air is less dense). At 1 atm, 0 degrees C, dry air is about 1.3 kg/m^3
v is velocity is meters/second
Cd is drag coefficient and depends on object. A fairly streamlined car is about 0.25, the open side of a hemisphere (or parachute) is about 1.4 (dimensionless)
A is the frontal area, in square meters.
With units above, force is in newtons. (English units would require a messy conversion factor)
A couple of links may offer some help:
http://www.windpower.org/en/tour/wtrb/drag.htm
http://damonrinard.com/aero/formulas.htm
Also, unless the object is completely symmerical about the wind vector, there is a lift force (at right angles to the wind) as well as a drag force. For some shapes (airplanes, by design) the lift force exceeds the drag force, and usually can't be neglected for "real" shapes.
Also the wind is never completely steady. The varying forces may invoke a resonance of the structure and analysis of this is very complex; even constant flow will generate Karman vortices off the edges. For simple shapes, frequency of Karman vortices can be estimated and may impose a larger oscillatory force than the "steady" force on the object, leading to structural failure. (Verrazano Narrows bridge).
Please don't presume to due structural analysis based on the simple formula above.