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3 years ago

The strength and stiffness of a properly constructed composite buildup depends primarily on?

Engineering

1 answer:

Flauer [41]3 years ago

4 0

Answer:orientation of piles

Explanation:

The stiffness and strength of a properly constructed composite buildup depend upon the orientation of piles to the load direction.

Structural composite materials are designed and manufactured to withstand certain specific stress loads. The ability to withstand these stress loads depends, to a large extent, on how the fibers are arranged. The fiber orientation of the parts must be the same as that of the original form.

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Mrrafil [7]

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3 years ago

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3 years ago

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What is thermodynamics system? Briefly explain (i) control mass system (ii) control volume system. Comment briefly on control ma

olga2289 [7]

Explanation:

Thermodynamics system :

Thermodynamics system is a region or space in which study of matters can be done.The system is separated from surroundings by a boundary this boundary maybe flexible or fixed it depends on situations.The out side the system is called surroundings.

Generally thermodynamics systems are of three types

1.Closed system(control mass system)

Only energy transfer take place ,no mass transfer take place.

2.Open system(control volume system)

Both mass as well as energy transfer take place.

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3 years ago

A rectangular steel bar, with 8" x 0.75" cross-sectional dimensions, has equal and opposite moments applied to its ends.

denpristay [2]

Answer:

Part a: The yield moment is 400 k.in.

Part b: The strain is $8.621 \times 10^{-4} in/in$

Part c: The plastic moment is 600 ksi.

Explanation:

Part a:

As per bending equation

$\frac{M}{I}=\frac{F}{y}$

Here

M is the moment which is to be calculated
I is the moment of inertia given as

$I=\frac{bd^3}{12}$

Here

b is the breath given as 0.75"
d is the depth which is given as 8"

$I=\frac{bd^3}{12}\\I=\frac{0.75\times 8^3}{12}\\I=32 in^4$

y is given as

$y=\frac{d}{2}\\y=\frac{8}{2}\\y=4"\\$

Force is 50 ksi

$\frac{M_y}{I}=\frac{F_y}{y}\\M_y=\frac{F_y}{y}{I}\\M_y=\frac{50}{4}{32}\\M_y=400 k. in$

The yield moment is 400 k.in.

Part b:

The strain is given as

$Strain=\frac{Stress}{Elastic Modulus}$

The stress at the station 2" down from the top is estimated by ratio of triangles as

$F_{2"}=\frac{F_y}{y}\times 2"\\F_{2"}=\frac{50 ksi}{4"}\times 2"\\F_{2"}=25 ksi$

Now the steel has the elastic modulus of E=29000 ksi

$Strain=\frac{Stress}{Elastic Modulus}\\Strain=\frac{F_{2"}}{E}\\Strain=\frac{25}{29000}\\Strain=8.621 \times 10^{-4} in/in$

So the strain is $8.621 \times 10^{-4} in/in$

Part c:

For a rectangular shape the shape factor is given as 1.5.

Now the plastic moment is given as

$shape\, factor=\frac{Plastic\, Moment}{Yield\, Moment}\\{Plastic\, Moment}=shape\, factor\times {Yield\, Moment}\\{Plastic\, Moment}=1.5\times400 ksi\\{Plastic\, Moment}=600 ksi$

The plastic moment is 600 ksi.

3 0

4 years ago