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harkovskaia [24]
3 years ago
13

A solid square rod is cantilevered at one end. The rod is 0.6 m long and supports a completely reversing transverse load at the

other end of +/- 2 kN. The material is AISI 1080 hot-rolled steel. If the rod must support this load for 104 cycles with a factor of safety of 1.5, what dimension should the square cross section have? Neglect any stress concentrations at the support end. Hint: Initially, assume the size endurance limit modification factor is kb=0.85; then check this assumption after the actual size is determined.
Engineering
1 answer:
Gnoma [55]3 years ago
3 0

Answer / Explanation:

Given the data :

Length of the rod (L) = 0.6m

Load;  F  =  ±  2  KN

Ultimate strength;  S u t = 770 M P a  = 770 / 6.89 kpsi = 112 kpsi

Yield strength;  S y = 420 M P a

Design factor = 1.5

Surface factor;  k a  =  0.488

Size factor;  k b  =  0.85

Now by using mechanical engineering design data book,

We calculate for fatigue strength

So, Se¹ = 0.5  ×  770

Se¹ = 385  M P a

K a =  57.7  ×  770⁻⁰⁷¹⁸

Kₐ = 0.488

Assume Kb = 0.85  (check later)

Se = Kₐ Kb Se¹

Se = 0.488  ×  0.85  ×  385

S e  =  160  M P a

By using design data book,

a = ( f Sut )² / Se

b = - 1 / 3 log f Sut / Se

S f  =  ₐ N ᵇ

Sf = 2553  (10⁴) ⁻ ⁰.²⁰⁰⁵

S f  =  403  M P a

The expression for stress,

M max =  2000  ×  0.6

M max  =  1200  Newton per meter

= M ( b/2) / I = M ( b/2) / bb³/12

= 6M / b³ = 6 x 1200 / b³

= 7200 / b³ Pa

Now by comparing strength and stress,

S f  =  n  ×  σ a

= 403  ×  10 ⁶

=  1.5  ×  7200/ b³

where b = 0.0299  m

           b = 30  m m

So, we assume the size factor, so the expression for checking the size factor is,

By using design data book,

de = 0.808  (hb)∧ 1/2

de = 0.808 b

de = 0.808  ×  30

de = 24.2

Kb = ( 24.2 / 7.62) ∧-0.107

Kb = 0.88

We have assume 0.85 which is nearer to the 0.88 and it is acceptable.

So the dimension of the square is 30 mm

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The ABC Corporation manufactures and sells two products: T1 and T2. 20XX budget for the company is given below:
Sliva [168]

Answer:

The ABC Corporation

a) Total Expected Revenue (in dollars) for 20XX:

Revenue from T1 = 60,000 x $165 = $26,400,000

Revenue from T2 = 40,000 x $250 = $10,000,000

Total Revenue from T1 and T2 = $36,400,000

b) Production Level (in units) for T1 and T2

                                           T1                       T2

Total Units sold             160,000           40,000

Add Closing Inventory   25,000             9,000

Units Available for sale 185,000           49,000

less opening inventory  20,000             8,000

Production Level          165,000 units 41,000 units

c) Total Direct Material Purchases (in dollars):

Cost of direct materials used    T1                T2

A:       (165,000 x 4 x $12)   $7,920,000   $2,460,000 (41,000 x 5 x $12)

B:       (165,000 x 2 x $5)       1,650,000          615,000 (41,000 x 3 x $5)

C:                                                           0          123,000 (41,000 x 1 x$3)

Total cost                            $9,570,000     $3,198,000 Total = $12,768,000

Cost of direct per unit = $58 ($9,570,000/165,000) for T1 and $78 ($3,198,000/41,000) for T2

Cost of direct materials used for production $12,768,000

Cost of closing direct materials:

                 A  (36,000 x $12)  $432,000

                 B (32,000 x $5)        160,000

                 C (7,000 x $3)            21,000             $613,000

Cost of direct materials available for prodn   $13,381,000

Less cost of beginning direct materials:

                 A  (32,000 x $12)        $384,000

                 B  (29,000 x $5)            145,000

                 C  (6,000 x $3)                18,000        $547,000

Cost of direct materials purchases               $12,834,000

d) The Total Direct Manufacturing Labor Cost (in dollars):

                                             T1                         T2

Direct labor per unit              2 hours                  3 hours

Direct labor rate per hour    $12                        $16

Direct labor cost per unit   $24                          $48

Production level              165,000 units        41,000 units

Labor Cost ($)                $3,960,000        $1,968,000

Total labor cost  $5,928,000 ($3,960,000 + $1,968,000)

e) Total Overhead cost (in dollars):

Overhead rate  = $20 per labor hour

Overhead cost per unit: T1 = $40 ($20 x 2) and T2 = $60 ($20 x 3)

T1 overhead = $20 x 2  x 165,000) = $6,600,000

T2 overhead = $20 x 3 x 41,000) =    $2,460,000

Total Overhead cost =                        $9,060,000

Cost of goods produced:

Cost of opening inventory of materials  = $547,000

Purchases of directials materials             12,834,000

less closing inventory of materials     =      $613,000

Cost of materials used for production    12,768,000

add Labor cost                                           5,928,000

add Overhead cost                                    9,060,000

Total production cost                            $27,756,000

f) Total cost of goods sold (in dollars):

Cost of opening inventory =          $3,928,000

Total Production cost             =    $27,756,000

Cost of goods available for sale  $31,684,000

Less cost of closing inventory       $4,724,000

Total cost of goods sold            $26,960,000

g) Total expected operating income (in dollars)

Sales Revenue:  T1 and T2  $36,400,000

Cost of goods sold                 26,960,000

Gross profit                             $9,440,000

less marketing & distribution      400,000

Total Expected Operating Income = $9,040,000

Explanation:

a) Cost of beginning inventory of finished goods:

T1, (Direct materials + Labor + Overhead) X inventory units =

T1 = 20,000 x ($58 + 24 + 40) = $2,440,000

T2 = 8,000 ($78 + 48 + 60) = $1,488,000

Total cost of beginning inventory = $3,928,000

b) Cost of closing Inventory of finished goods:

T1 = 25,000 x ($58 + 24 + 40) = $3,050,000

T2 = 9,000 ($78 + 48 + 60) = $1,674,000

Total cost of closing inventory = $4,724,000

5 0
4 years ago
A hole of diameter D = 0.25 m is drilled through the center of a solid block of square cross section with w = 1 m on a side. The
attashe74 [19]

Answer:

q=4.013\:\:kW\\\\T_1=278.91\:\:^{\circ}C\\\\T_2=275.82\:\:^{\circ}C

Explanation:

R_{conv,1}=(h_1\pi D_1L)^{-1}=(50*0.25*2)^{-1}=0.01273\:\:K/W\\\\R_{conv,2}=(h^2*4wL)^{-1}=(4*4*1)^{-1}=0.0625\:\:K/W\\\\R_{cond(2D)}=(Sk)^{-1}=(8.59*150)^{-1}=0.00078\:\:K/W

So, heat rate can be calculated as follows:

q=\frac{T_{\infty,1}-T_{\infty,2}}{R_{conv,1}+R_{conv,2}+R_{cond(2D)}} =\frac{330-25}{0.076} =4.013\:\:kW

Surface temperatures can be calculated as follows:

T_1=T_{\infty,1}-qR_{conv,1}=330-51.09=278.91\:\:^{\circ}C\\\\T_2=T_{\infty,2}+qR_{conv,2}=25+250.82=275.82\:\:^{\circ}C

6 0
4 years ago
Air initially at 120 psia and 500o F is expanded by an adiabatic turbine to 15 psia and 200o F. Assuming air can be treated as a
Goshia [24]

Answer:

a. Wa = 73.14 Btu/lbm

b. Sgen = 0.05042 Btu/lbm °R

c. Isentropic efficiency is 70.76%

d. Minimum specific work for compressor W = -146.2698 Btu/lbm [It is negative because work is being done on the compressor]

Explanation:

Complete question is as follows;

Air initially at 120 psia and 500oF is expanded by an adiabatic turbine to 15 psia and 200oF. Assuming air can be treated as an ideal gas and has variable specific heat.

a) Determine the specific work output of the actual turbine (Btu/lbm).

b) Determine the amount of specific entropy generation during the irreversible process (Btu/lbm R).

c) Determine the isentropic efficiency of this turbine (%).

d) Suppose the turbine now operates as an ideal compressor (reversible and adiabatic) where the initial pressure is 15 psia, the initial temperature is 200 oF, and the ideal exit state is 120 psia. What is the minimum specific work the compressor will be required to operate (Btu/lbm)?

solution;

Please check attachment for complete solution and step by step explanation

8 0
4 years ago
A clay sample has a wet mass of 417 g, a volume of 276 cm3, and a specific gravity of 2.70. When oven dried, the mass becomes 22
nataly862011 [7]

Answer:

a. WATER CONTENT.

mass of water (MW)= 417g - 225g

Mw= 192g

M= (Mw / Ms)

M= [192/417]

M= 0.46

M= 46.0%

b. Void Ration (e)

To calculate the void ratio we must first calculate the volume of solids. Then we can find the volume of voids by subtracting the volume of solids from the total volume.

Ps= (Ms/Vs)=GsPw

Therefore

Vs= (Ms/GsPw)

Vs= (417/ 2.70*1000)

Vs= 0.154cm3

But V= Vv+Vs

Vv=V-Vs

e= (Vv/Vs)

8 0
3 years ago
What is the purpose of a weld symbol
andreev551 [17]

Answer:

To indicate the welding or cutting processes

8 0
3 years ago
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