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2 years ago

You guys want to talk at seven I am free then

Engineering

2 answers:

Viefleur [7K]2 years ago

8 0

Answer:

Hello my friend how are you

myrzilka [38]2 years ago

7 0

Answer:

sure? lol (btw can I get brainliest?)

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Bind hole, 38 diameter, .50 deep

agasfer [191]

Answer:

59.69021

Explanation:

38/.5 x 3.14159

4 0

2 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

3 years ago

At steady state, air at 200 kPa, 325 K, and mass flow rate

Vera_Pavlovna [14]

Letra A

A letra

A.
Thank

3 0

3 years ago

The interior wall of a building is made from 2×4 wood studs, plastered on one side. If the wall is 13 ft high, determine the loa

Elanso [62]

Answer:

load = 156 lb/ft

Explanation:

given data

interior wall of a building = 2×4 wood studs

plastered = 1 side

wall height = 13 ft

solution

we get here load so first we get wood stud load and that is

we know here from ASCE-7 norm

dead load of 2 x 4 wood studs with 1 side plaster = 12 psf

and we have given height 13 ft

so load will be = 12 psf × 13 ft

load = 156 lb/ft

7 0

3 years ago

A plate (A-C) is connected to steelflat bars by pinsat A and B. Member A-E consists of two 6mm by 25mm parallel flat bars. At C,

juin [17]

Answer:

stress_ac = 5.333 MPa

shear stress_c = 1.763 MPa

Explanation:

Given:

- The missing figure is in the attachment.

- The dimensions of member AC = ( 6 x 25 ) mm x 2

- The diameter of the pin d = 19 mm

- Load at point A is P = 2 kN

Find:

- Find the axial stress in AE and the shear stress in pin C.

Solution:

- The stress in member AE can be calculated using component of force P along the member AE as follows:

stress_ac = P*cos(Q) / A_ae

Where, Angle Q: A_E_B and A_ac: cross sectional area of member AE.

cos(Q) = 4 / 5 ..... From figure ( trigonometry )

A_ae = 0.006*0.025*2 = 3*10^-4 m^2

Hence,

stress_ae = 2*(4/5) / 3*10^-4

stress_ae = 5.333 MPa

- The force at pin C can be evaluated by taking moments about C equal zero:

(M)_c = P*6 - F_eb*3

0 = P*6 - F_eb*3

F_eb = 0.5*P

- Sum of horizontal forces for member AC is zero:

P - F_eb - F_c = 0

F_c = 0.5*P

- The shear stress of double shear bolt is given by an expression:

shear stress = shear force / 2*A_pin

Where, The area of the pin C is:

A_pin = pi*d^2 / 4

A_pin = pi*0.019^2 / 4 = 2.8353*10^-4 m^2

Hence,

shear stress = 0.5*P / 2*A_pin

shear stress = 0.5*2 / 2*2.8353*10^-4

shear stress = 1.763 MPa

7 0

3 years ago