Construct the plane-stress yield envelopes in a principle stress space for both the Tresca and the von Mises yield theories using your calculated value of the yield strength to scale the envelopes. Indicate the two equivalent load paths corresponding to pure shear on the yield envelopes. Calculate the shear yield strength of Al 6061-T6 aluminum predicted by the above theories.

Answer:

it is indeed C

Explanation:

Answer:

c

Explanation:

Answer:

C Electrical Connections

Explanation:

In reading says . However, electrical

connections aren’t shown in construction drawings.

Answer: hello your question lacks the required diagram attached below is the diagram

answer :  29528.1  N/m^2

Explanation:

Given data :

dimensions of tank :

Length = 5-m

Width = 4-m

Depth = 2.5-m

acceleration of tank = 2m/s^2

Determine the maximum gage pressure in the tank

Pa ( pressure at point A )  = s*g*h1

    = 10^3 * 9.81 * 3.01

    = 29528.1  N/m^2

attached below is the remaining part of the solution

Answer:

a) The maximum flowrate of the pump is approximately 13,305.22 cm³/s

b) The pressure difference across the pump is approximately 293.118 kPa

Explanation:

The efficiency of the pump = 78%

The power of the pump = 5 -kW

The height of the pool above the underground water, h = 30 m

The diameter of the pipe on the intake side = 7 cm

The diameter of the pipe on the discharge side = 5 cm

a) The maximum flowrate of the pump is given as follows;

[tex]P = \dfrac{Q \cdot \rho \cdot g\cdot h}{\eta_t}[/tex]

Where;

P = The power of the pump

Q = The flowrate of the pump

ρ = The density of the fluid = 997 kg/m³

h = The head of the pump = 30 m

g = The acceleration due to gravity ≈ 9.8 m/s²

[tex]\eta_t[/tex] = The efficiency of the pump = 78%

[tex]\therefore Q_{max} = \dfrac{P \cdot \eta_t}{\rho \cdot g\cdot h}[/tex]

[tex]Q_{max}[/tex] = 5,000 × 0.78/(997 × 9.8 × 30) ≈ 0.0133 m³/s

The maximum flowrate of the pump [tex]Q_{max}[/tex] ≈ 0.013305 m³/s = 13,305.22 cm³/s

b) The pressure difference across the pump, ΔP = ρ·g·h

∴ ΔP = 997 kg/m³ × 9.8 m/s² × 30 m = 293.118 kPa

The pressure difference across the pump, ΔP ≈ 293.118 kPa

Answer:

c   Waste stack

Explanation:

Answer:

a. 2.30

b. decreases with increasing velocity.

c. 0.179 kg/s.

Explanation:

Without mincing let's dive straight into the solution to the question above.

                                                         [a].

The improvement in cooling rate that can be achieved using the slotted nozzle arrangement in lieu of turbulated air at 10 m/s and 15°C in parallel flow over the plate can be determined by calculating turbulent flow:

The turbulent flow over the plate= 10 × 0.5/ 20.92 × 10⁻6 = 2.39 × 10⁵.

While the turbulent flow correlation = 0.037( 2.39 × 10⁵)^[tex]\frac{4}{5}[/tex] (0.7)^[tex]\frac{1}{3}[/tex] = 659.6.

Array of slot noozle = [10 × (2  × 0.004)]/ 20.92  × 10^-6] = 3824.

where A = 4/56 =0.714.

And Ar = [ 60 + 4 (40/2  × 4) - 2 ]^2 ]-1/2 = 0.1021.

N = 2/3 (0.1021)^3/4 [ 2  ×  3824/ ( 0.0714 / 0.1021) + (.1021/0.0714)] (0.700)^0.42 =24.3.

h = 24.3  ×  0.030/0.004 = 91.1 W/m^2k.

Therefore; 659.6  × 0.030/0.5 = 39.0 W/m²k.

The turbulent flow = 0.5 × 39.6 × 0.5( 140 -15) = 1237.5 W.

The slot noozle = 91.1  ×  0.5  ×  0.5 [ 140 -15] = 2846.87W.

The improvement in cooling rate = 2846.87/ 1237.5 = 2.30.

                                                     [b].

2.3 [ (2^2/3)/ 2^4/5] = 2.1

Thus, it decreases with increasing velocity

                                                      [c].

The  air mass rate requirement for the slotted nozzle arrangement = 9 × 0.995 (0.5 × 0.004)10 = 0.179 kg/s.

Answer:

shaft diameter = [tex]\sqrt[3]{0.3512}[/tex] mm = 0.7055 mm

Explanation:

Ratio of inside diameter to outside diameter ( i.e. d/D )= 0.6

Shear stress of material ( Z ) ≤ 500 KPa

power transmitted by shaft ( P ) = 1.5MW of mechanical power

Revolution ( N ) = 1500 rev/min

Calculate shaft Diameter

Given that: P = [tex]\frac{2\pi NT}{60}[/tex]  ---- 1

therefore; T = ( 1.5 *10^3 * 60 ) / ( 2[tex]\pi[/tex] * 1500 )  = 9.554 KN-M

next

[tex]\frac{T}{I_{p} } = \frac{Z}{R}[/tex]

hence ; T = Z[tex]_{p} *Z[/tex]

attached below is the remaining part of the solution

Answer:

10.007

Explanation:

Assuming we have to find out the compression ratio of the engine

Given information

Cubic capacity of the engine, V = 245 cc

Clearance volume, V_c = 27.2 cc

over square-ratio = 1.1

thus,

D/L = 1.1

where,

D is the bore

L is the stroke

Now,

Volume of the engine V =[tex]\frac{\pi}{4} D^2L[/tex]

plugging values we get

245 = [tex]\frac{\pi}{4} D^3/1.1[/tex]

Solving we get D =7 cm

therefore,  L= 7/1.1 =6.36 cm

Now,

the compression ratio is given as:

r =(V+V_c)/V_c

on substituting the values, we get

r = (245+27.2)/27.2 =10.007

Hence, Compression ratio = 10.007

Answer:

13.4 mm

Explanation:

Given data :

Load amplitude ( F )  = 22,000 N

factor of safety ( N )= 2.0

Take ( Fatigue limit stress amplitude for this alloy ) б = 310 MPa

calculate the minimum allowable bar diameter to ensure that fatigue failure will not occur

minimum allowable bar diameter = 13.4 * 10^-3 m ≈ 13.4 mm

attached below is a detailed solution

Answer:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.

Explanation:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers. ... This means that you need good wholesalers and good transportation systems to get your products to the retailers.

Answer:

Distribution (or its more sophisticated counterpart, supply chain management) can add value to goods and services by making them more easily and conveniently available to consumers.

Explanation:

hope it helps <33

Answer:

Explanation:

Crystalline melting temperature (Tm) is termed as the temperature required for a  crystalline polymer to change to a fluid or glasslike crystalline spaces of a semi-crystalline polymer liquefy (expanded sub-atomic movement).  

Crystallization of polymers is an interaction related with incomplete arrangement of their atomic and molecular chains. These chains crease together and structure requested districts called lamellae, which form bigger spheroidal designs named spherulites. Polymers can solidify after cooling from melting, mechanical extending, or dissolvable dissipation. Crystallization influences the optical, mechanical, and synthetic chemical properties of the polymer.  

For a crystalline polymer, a required polymer chain is present in or goes along a few crystalline and amorphous zones. The crystalline zones are comprised of intermolecular & intramolecular arrangements or deliberate and thus firmly stuffed plan of atoms or chain fragments, and an absence of it brings about the development of amorphous zones.  

The mechanical property boundary, for example, shear modulus expansions in the temperature of perception for polymer material framework.  

The temperature reaction of direct linear polymers might be seen as partitioned into three particularly separate fragments:  

1. Above Tm: The polymer stays as fluid whose consistency & viscosity would rely upon atomic molecular weight and temperature.  

2. Between Tm and Tg: This area may go between close to 100% crystalline & 100% amorphous chain atomic bunches relying upon the polymer underlying consistency. The amorphous part carries on similar to supercooled fluid in this section. The generally actual conduct of the polymer in this moderate portion is similar to an elastic rubber.  

3.Below Tg: The polymer material saw as glass is hard and inflexible, showing and emanating a predetermined coefficient of thermal extension. The glass is more like a crystalline strong than the fluid in personal conduct standard regarding mechanical property boundaries. In regard to the molecular atomic request, in any case, the glass all the more intently takes after the fluid. There is little contrast between the direct linear and cross-connected polymer beneath Tg.