Question

The mat foundation on Figure #3 supports four silos. Each silo has an empty weight of 7000 kips and can hold up to 20,00 kips of corn. The mat has a weight of 13,500 kips. Since each silo is filled independently, evaluate various loading conditions and determine whether eccentric loading requirements will be met. If these requirements are not met, determine the minimum mat width, B, to satisfy these requirements. (35 Points) 100 FT 23 FT 23 FT 100 FT

Answer 1

Given,

Dead load = 7000 kips

Take diameter of silo is 40 ft. = 12.192 m

And it's height is 3 x diameter. = 120 ft = 36.576 m

Area = $\pi$ /4 x d²

= 116.74 m²

Live load = 20000 kips

Total load = 27000 kips

= 120101.98 KN

Impact loads = + 50% of live load

= 20000 kips x 1.5

= 30000 kips

= 133446.3 KN

Snow load = 75 Kg/m²

Ls = 75 x 116.74 m²

= 85.89 KN

Let wind velocity V be 14 m/s

Wind load = 0.6 V²/10

= 0.6 x 14² /10

= 11.76 Kg/m²

Lw = 11.76 Kg/m² x 116.74 m²

= 13.467 KN

Weight of mat given = 13500 kips

Total weight of 4 silos = 27000 kips x 4

= 108000 kips

Total weight of structure W = 108000 kips + 13500 kips

= 121500 kips

Earthquake force, F = W x a/g

Earthquake acceleration, a = 0.2 m/s²

Acceleration of gravity, g = 9.81 m/s²

$\therefore$ F = 121500 kips x 0.2/9.81

= 2477.06 kips

Now, Moment of silo is 27000 kips x height

M = 120101.98 KN x 36.576 m

= 4392847 KNm

So, eccentricity is M/ Total forces on 1 silo

= 4392847 KNm / ( 120101.98 KN + 133446.3 KN + 85.89 KN + 13.467 KN)

= 17.31 m

It is much safer to space each silo at least 3 x diameter from each center.

Increment of length = 40 x 4 = 160 ft minimum.

B ~ 200 ft x 200 ft.