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andre [41]
3 years ago
8

Chapter 15 – Fasteners: Determine the tensile load capacity of a 5/16 – 18 UNC thread and a 5/16 – 24 UNF thread made of the sam

e material. Which is stronger?
Make the same comparison for M8 x 1.25 and M8 x 1 ISO threads. Hint: see Tables 15-1, 15-2, and 15-3
Engineering
2 answers:
Roman55 [17]3 years ago
7 0

Answer:

The 5/16 – 24 UNF is stronger because it has more tensile load capacity.

Tensile load capacity for M8 -1.25 = 5670 lb

Tensile load capacity for M8 -1 = 6067 lb

Explanation:

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 5/16 – 24 UNF is stronger because it has more tensile load capacity.

For metric Bolts:

We have to consider all values in SI units

Strength = 689 MPa

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 = 6067lb

Aliun [14]3 years ago
7 0

Answer:

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

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 = 6067N

Strength of 5/16 - 14 Acme thread

We use the imperial fomula as mentioned above,

The tensile capacity is 4634 lb.

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4 years ago
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Four race cars are traveling on a 2.5-mile tri-oval track. The four cars are traveling at constant speeds of 195 mi/h, 190 mi/h,
Snezhnost [94]

Answer:

Explanation:

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

1) The number of times, the car with the speed of  195 mph will cross the given point is equal to 30 minutes divided by the time taken by car to cross the 2.5 miles.

0 .5*195/2.5 = 39

Likewise, the car with the speed of 190 mph crosses the point 38 times; the car with the speed of 185 mph crosses the point 37 times

and car with the speed of 180 mph crosses it 36 times

here, the time-mean speed, vt is given below,

vt = (39*195 +38*190+37*185+36*180)/(39+38+37+38)

= 186.433 mph

and space mean speed is given by,

= (39+38+37+36)/(39/195+38/190+37/1850+36/180)

=187.5 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

2)  There would be only four number of observations when the aerial photo is given, therefore time mean speed, vt, in that condition will be calculated as

Vt = 195+190+185+180/4

  = 187.5

Vs= 4/(1/195+1/190+1/185+1/180)

= 188.36 mph

4 0
4 years ago
What is the primary responsibility of ABET?
nalin [4]

Answer:

C

Explanation:

The ABET (Accreditation Board for Engineering and Technology) is a non-governmental organization that accredits programs in applied science, computing, engineering, and engineering technology, both in the United States and elsewhere.

Give Brainliest pls

5 0
3 years ago
A water tank is completely filled with liquid waterat 20°C.The tank material is such that it can withstand tensioncaused by a vo
Xelga [282]

Answer:

Highest temperature rise allowable = ΔT = 21.22°C

Highest allowable temperature = ΔT + 20 = 41.22°C

Explanation:

From literature, the coefficient of volume expansion of water between 20°C and 50°C = β = (0.377 × 10⁻³) K⁻¹

Volume expansivity is given by

ΔV = V β ΔT

ΔV = Change in volume

V = initial volume

β = Coefficient of volume expansion = (0.377 × 10⁻³) K⁻¹ = 0.000377 K⁻¹

ΔT = Change in temperature = ?

It is given in the question that maximum volume increase the tank can withstand is

(ΔV/V) × 100% = 0.8%

(ΔV/V) = 0.008

V β ΔT = ΔV

β ΔT = (ΔV/V)

β ΔT = 0.008

ΔT = (0.008/β)

ΔT = (0.008/0.000377)

ΔT = 21.22°C

Highest temperature rise allowable = ΔT = 21.22°C

Highest allowable temperature = ΔT + 20 = 41.22°C

Hope this Helps

5 0
4 years ago
An overhead 25m-long, uninsulated industrial steam pipe of 100-mm diameter, is routed through a building whose walls and air are
wariber [46]

Answer:

1) q=18414.93 W

2) C=12920$

Explanation:

Given data:

pipe length L=25m

pipe diameter D=100mm =0.1 m

air temperature T_{s1}=T_{\infty1} }=25 °C.....= 298.15k

pipe surface temp T_{s2}=150 °C.....=423.15k

surface emissivity e= 0.8

boiler efficiency η=0.90

natural gas price Cg=$0.02 per MJ

1) Total heat loss and radiation heat loss combined

          q=q_{conv} +q_{rad}

          q=A[h(T_{s2}-T_{s1})+eб(T_{s2}^4-T_{s1}^4)]....... (1)

б=5.67×10^-8 W/m^2K^4 (boltzmann constant)

area A =L.Dπ=25×0.1π=7.85 m^2

putting all these values in eq (1)

q=18414.93 W

2) suppose boiler is operating non stop annual energy loss will be

               E=q.t

                  =18414.93.3600.24.365

                  =5.81×10^11 J

   to find furnace energy consumption

               Ef =E/η

                  =6.46×10^5 MJ

   annual cost

                  C=Cg. Ef

                    =12920$

8 0
3 years ago
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