A. How wide must a wall footing be if the load is 9,500 pounds per foot of wall length, and the footing rests on a sandy gravel? Show calculations and make a sketch. Assume that it is 18" thick. b. How large must a square column footing be to support a load of 65,000 pounds if it stands on silty gravel? Show calculations. Make a sketch of the footing, assuming that it is 18" thick. © Diane Reid

a) 38"

b) L = 5.7'

Step-by-step explanation:

a) Given

P = 9,500 lb/ft

Thick = 18"

We assume an allowable bearing pressure of σ = 3000 lb/sf

If Stress = Force / Area

σ = P/A

Solving for area

A = P/σ  ⇒  A = 9,500 lb/ft/3000 lb/sf

⇒  A = 3.167 sf,  or  3.167 ft wide, per foot of length.

⇒  A = 38" wide

We can see the pic 1 in order to understand the answer.

b) Given

P = 65,000 pounds = 65,000 lb

Thick = 18" = 1.5'

We assume an allowable bearing pressure of σ = 2000 lb/sf

If Stress = Force / Area

σ = P/A

Solving for area

A = P/σ  ⇒  A = 65,000 lb/2000 lb/sf

⇒  A = 32.5 sf

then A = L²  ⇒  L = √A = √(32.5 sf) = 5.7 ft

Finally L = 5.7'

We can see the pic 2 in order to understand the answer.

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step-by-step explanation:

1+2+3+4+5+6+7+8+9+10 = 55

2*10 = 20

55-20 = 35

hello from mrbilldoesmath!

answer:

the domain is the set { -2, 0, 2, 3}

discussion:

the domain is the set of x values in the ordered pairs (x,y). taking the x, or first component, of each ordered pair gives   -2, 0, 2, 3.

you,

mrb