Answer:
Explanation:
The Young's module is:
Let assume that both specimens have the same geometry and load rate. Then:
The displacement rate for steel is:
Using an "AND" and an "OR", list all information (Equipment Number, Equipment Type, Seat Capacity, Fuel Capacity, and Miles per
Gallon) on aircraft that have a seat capacity less than 250, or aircraft that have a miles per gallon greater than 4.0 miles per gallon and fuel capacity less than 2500.
1 answer:
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Answer:
correct option is (B) 315,500
Explanation:
given data
volume = 320,000 yd³ = 8640000 ft³
dry unit weight = 106 pcf
moisture content = 18.2%
total unit weight = 122 pcf
moisture content = 16.7%
to find out
volume of borrow lyd needed
solution
first we get here weight of material that is
weight = volume × unit weight
weight = 8640000 × 122
weight = 1054080000 lb
that weight is weight of water + weight of solid so
0.167 × weight is weight of water + weight of solid ) = 1054080000 lb
and weight of solid =
weight of soil solid is = 903239075 pound
and weight of water = 150840925 pound
so volume of soil = 903239075 ÷ 106 lb/ft³ = 8521123.34 ft³
and volume required = 8521123.34 ft³ ÷ 27 ft³ = 315597.161 yd³
volume required = 315500 yd³
so correct option is (B) 315,500
Answer:
critical stress = 1382.67 MPa
Explanation:
given data
plane strain fracture toughness = 54.8 MP
length of surface creak = 0.5 mm
we take here
parameter Y = 1.0
solution
we apply critical stress formula that is
critical stress =
.............................1
here K is design stress plane strain fracture toughness and a is length of surface creak so put all these value in equation 1
critical stress =
solve it we get
critical stress = 1382.67 MPa
As exposed stress 1030 MPa is less than critical stress 1382 MPa
so that fracture will not be occur here