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madreJ [45]
3 years ago
6

Compare the tensile load capacity of a 5/16-18 UNC thread and a 5/16-24 UNF thread made of the same material.

Engineering
1 answer:
ehidna [41]3 years ago
6 0

Answer:

Explanation:

The material strength is given as 100,000 PSI  

To find the tensile load capacity of a threaded rod, we multiply the Strength of the material with the Stress area for the bolt.

A_{t} = 0.7854 (D -  {0.9743} / n))²

Where D is the Basic Major Diameter

n is the Number of threads per inch

At is the Tensile strength area of the bolt.

For 5/16 - 18 UNC thread

D = 0.3125

n = 18

Therefore the tensile load capacity is = 100000 X (0.7854 X (0.3125 - 0.9743/ 18)²

= 5243 lb.

Similarly for 5/16 - 24 UNF , only the n value changes to 24

we get the tensile load capacity = 5806.6 lb

Hence the fine thread bolt is stronger because it has more tensile load capacity.

For metric Bolts

We have to consider all values in SI units

Strength = 689 MPa

A_{t} = {4} / (D - 0.938194 X pitch)^{2}

We get for M8 -1.25

Tensile load capacity as = 689 X 36.6 = 25223 N = 5670 lb

For M8 -1

Tensile load capacity as = 689 X 39.167 = 26986 N = 6067 N

Strength of 5/16 - 14 Acme thread

We use the imperial formula as mentioned above,

The tensile capacity is 4634 lb.

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The waffle slab is: a) the two-way concrete joist framing system. b) a one-way floor and roof framing system. c) the one-way con
GREYUIT [131]

Answer:

a) the two-way concrete joist framing system

Explanation:

A waffle slab is also known as ribbed slab, it is a slab which as waffle like appearance with holes beneath. It is adopted in construction projects that has long length, length more than 12m. The waffle slab is rigid, therefore it is used in building that needs minimal vibration.

4 0
3 years ago
Ma poate ajuta cineva?
kari74 [83]
Da, sigur. cu ce ai nevoie de ajutor?
5 0
3 years ago
A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 5. Find the volume of the ring
ANTONII [103]

Answer:

The volume of the ring shaped solid that remains is 21 unit^3.

Explanation:

The total volume of the sphere is given as:

Volume of Sphere = (4/3)πr^3

where, r = radius of sphere

Volume of Sphere = (4/3)(π)(5)^3

Volume of Sphere = 523.6 unit^3

Now, we find the volume of sphere removed by the drill:

Volume removed = (Cross-sectional Area of drill)(Diameter of Sphere)

Volume removed = (πr²)(D)

where, r = radius of drill = 4

D = diameter of sphere = 2*5 = 10

Therefore,

Volume removed = (π)(4)²(10)

Volume removed = 502.6 unit^3

Therefore, the volume of ring shaped solid that remains will be the difference between the total volume of sphere, and the volume removed.

Volume of Ring = Volume of Sphere - Volume removed

Volume of Ring = 523.6 - 502.6

<u>Volume of Ring = 21 unit^3</u>

5 0
3 years ago
3. Briefly explain the conduction mechanism in metals?​
Bingel [31]

Answer:

conduction occurs when a substance is heated

Explanation:

3 0
2 years ago
Calculate the viscosity(dynamic) and kinematic viscosity of airwhen
nikitadnepr [17]

Answer:

(a) dynamic viscosity = 1.812\times 10^{-5}Pa-sec

(b) kinematic viscosity = 1.4732\times 10^{-5}m^2/sec

Explanation:

We have given temperature T = 288.15 K

Density d=1.23kg/m^3

According to Sutherland's Formula  dynamic viscosity is given by

{\mu} = {\mu}_0 \frac {T_0+C} {T + C} \left (\frac {T} {T_0} \right )^{3/2}, here

μ = dynamic viscosity in (Pa·s) at input temperature T,

\mu _0= reference viscosity in(Pa·s) at reference temperature T0,

T = input temperature in kelvin,

T_0 = reference temperature in kelvin,

C = Sutherland's constant for the gaseous material in question here C =120

\mu _0=4\pi \times 10^{-7}

T_0 = 291.15

\mu =4\pi \times 10^{-7}\times \frac{291.15+120}{285.15+120}\times \left ( \frac{288.15}{291.15} \right )^{\frac{3}{2}}=1.812\times 10^{-5}Pa-swhen T = 288.15 K

For kinematic viscosity :

\nu = \frac {\mu} {\rho}

kinemic\ viscosity=\frac{1.812\times 10^{-5}}{1.23}=1.4732\times 10^{-5}m^2/sec

3 0
3 years ago
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