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nadezda [96]
3 years ago
11

Three natural materials that were used by early man as roof covering​

Engineering
2 answers:
olga55 [171]3 years ago
8 0

Explanation:

animal parts, wood, rock, and clay

Law Incorporation [45]3 years ago
8 0

Answer:

roofing sheet

wood

metal

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For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm and the stiffness of the members is km = 2.
rjkz [21]

Answer:

a) 0.978

b) 0.9191

c) 1.056

d) 0.849

Explanation:

Given data :

Stiffness of each bolt = 1.0 MN/mm

Stiffness of the members = 2.6 MN/mm per bolt

Bolts are preloaded to 75% of proof strength

The bolts are M6 × 1 class 5.8 with rolled threads

Pmax =60 kN,  Pmin = 20kN

<u>a) Determine the yielding factor of safety</u>

n_{p} = \frac{S_{p}A_{t}  }{CP_{max}+ F_{i}  }  ------ ( 1 )

Sp = 380 MPa,   At = 20.1 mm^2,   C = 0.277,  Pmax = 7500 N,  Fi = 5728.5 N

Input the given values into the equation above

equation 1 becomes ( np ) = \frac{380*20.1}{0.277*7500*5728.5} = 0.978

note : values above are derived values whose solution are not basically part of the required solution hence they are not included

<u>b) Determine the overload factor of safety</u>

n_{L} =  \frac{S_{p}A_{t}-F_{i}   }{C(P_{max} )}  ------- ( 2 )

Sp =  380 MPa,   At =  20.1 mm^2, C = 0.277,  Pmax = 7500 N,  Fi = 5728.5 N

input values into equation 2 above

hence : n_{L} = 0.9191n_{L}  = 0.9191

<u>C)  Determine the factor of safety based on joint separation</u>

n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }

Fi =  5728.5 N,  Pmax = 7500 N,  C = 0.277,

input values into equation above

Hence n_{0} = 1.056

<u>D)  Determine the fatigue factor of safety using the Goodman criterion.</u>

nf = 0.849

attached below is the detailed solution .

4 0
3 years ago
Just need someone to talk to pls dont just use me for points
Allushta [10]

Answer:

well what do you wanna talk about friend?

Explanation:

7 0
3 years ago
Calculate the load, PP, that would cause AA to be displaced 0.01 inches to the right. The wires ABAB and ACAC are A36 steel and
Nataly [62]

Answer:

P = 4.745 kips

Explanation:

Given

ΔL = 0.01 in

E = 29000 KSI

D = 1/2 in  

LAB = LAC = L = 12 in

We get the area as follows

A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²

Then we use the formula

ΔL = P*L/(A*E)

For AB:

ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB

For AC:

ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)

⇒  ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC

Now, we use the condition

ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in

⇒  ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in

Knowing that   PAB*Cos 30°+PAC*Cos 30° = P

we have

(2.107*10⁻⁶ in/lbf)*P = 0.01 in

⇒  P = 4745.11 lb = 4.745 kips

The pic shown can help to understand the question.

5 0
3 years ago
What does STP and NTP stands for in temperature measurement?
Lisa [10]

STP stands for standard temperature pressure and NTP stands for normal temperature pressure

8 0
3 years ago
Mohr's circle represents: A Orientation dependence of normal and shear stresses at a point in mechanical members B The stress di
blsea [12.9K]

Answer:

The correct answer is A : Orientation dependence of normal and shear stresses at a point in mechanical members

Explanation:

Since we know that in a general element of any loaded object the normal and shearing stresses vary in the whole body which can be mathematically represented as

\sigma _{x'x'}=\frac{\sigma _{xx}+\sigma _{yy}}{2}+\frac{\sigma _{xx}-\sigma _{yy}}{2}cos(2\theta )+\tau _{xy}sin(2\theta )

And \tau _{x'x'}=-\frac{\sigma _{xx}-\sigma _{yy}}{2}sin(2\theta )+\tau _{xy}cos(2\theta )

Mohr's circle is the graphical representation of the variation represented by the above 2 formulae in the general oriented element of a body that is under stresses.

The Mohr circle is graphically displayed in the attached figure.

4 0
3 years ago
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