machine design

1. The lid of a cast iron Dutch oven pressure vessel is held in place by ten ½ in steel bolts having an initial

tightening load of 5000 lb, when the vessel is at 70 F and the initial pressure is atmospheric.

Determine the load in each bolt:

a. if the pressure is increased to 200 psi

b. if the vessel is heated to 250 F with atmospheric internal pressure

c. if the vessel is heated to 250 F with an internal pressure of 200 psi

Assume the following data.

Steel: E = 30106 psi, =6.610-6 / F

Cast Iron: E = 12106 psi, =5.610-6 / F

Zinc: 12106 psi, =17.810-6 / F

Ans./hints:

W* = 5360 lb, the new initial bolt load

c. After the external pressure has been applied W** = 5360+4020/37 = 5469 lb

Thus, W = 4020 lb / bolt & W’ = 5000+4020/37 = 5110 lb

2. A butt joint whose rivet pattern is shown below is to be used to form a circular steel boiler. The boiler

is to have a 5 ft diameter and be able to withstand an internal pressure of 500 psi.

Determine the required main plate thickness as well as the efficiency of the joint. The rivets are to be

7/8 inches dia., and the rivet holes are drilled to 15/16 inches dia. The working stresses are as follows:

Tension = 20,000 psi

Shear = 16,000 psi

Bearing = 20,000 psi in single shear

Bearing = 28,000 psi in double shear

Ans./hints:

Rivets are in double shear, P = nSA, P = 212,000 lb Tensile load acting on the boiler plates, 2T = p2RL, L = T/pR = 14.1 in Determine thickness of main plate: Pallow = nSallow d*tplate tplate = 0.79 in Tension row D: Pallow = Sallow(w-d)tplate tplate = 0.81 in Joint Efficiency = Pallow / Stensile, plate=212,000 lb/(20,000 psi*14.1 in *0.81 in)=93%

3. A Belleville spring is required to provide a constant force of 200 10 N over a deflection of 0.3 mm.

The spring must fit within a 62 mm dia. hole. A carbon spring steel with UTS = 1700 MPa is being

proposed. Recall,

And the thickness required to accomplish a particular force in the flat position (E = 207 GPa, Poison’s

Ratio =0.3, K1 = 0.69),

0.25 2

1

1072 /

o flatD F t

h t

    

   

Determine:

a. t (in)

b. h (in)

c. max (in)

d. min (in)

e. K1, K2, K3, K4, K5

f. c (MPa)

g. ti (MPa)

h. to (MPa)

Ans. h = t = 1.414 mm, min./max. deflections = 0.724 mm, 1.504 mm, K1 = 0.689, K2 = 1.220, K3 = 1.378,

K4 = 1.115, K5 = 1,c = -840 MPa, ti =355 MPa, to =658 MPa

4. Determine the required number of coils and permissible deflection in a helical spring fabricated of

1/16 inch dia. steel wire, assuming a spring index of 6 and an allowable stress of 50,000 psi in shear. The

spring rate is to be 10 lb/in.

Ans.

D = 0.375 in, Wahl factor: K = 1.25, n = 41.6 turns, F = 10.2 lb, = 1 in  k = 10 lb/in

5. One helical spring is nested inside another, the dimensions are as given:

Both springs have the same free length and carry a total maximum load of 550 lb. Determine:

a. the maximum load carried by each spring

b. the total deflection of each spring

c. the maximum stress in the outer spring

Assume G = 12106 psi.

Ans.

a. Fi = 482 lb, Fo = 68 lb

b. =1.325 in

c. C = 7, Wahl Factor: K = 1.213, Max. stress in outer spring = 41,700 psi

6. The load on an oil-tempered carbon helical compression spring varies from 150 lb to 400 lb. The mean

dia. of the coil is to be 2 inches, and the desired design factor of safety is F.S. = 1.3 based on variable

stresses. Determine the required wire size. Assume allowable yield stress in shear = 100,000 psi.

Ans. No. 4-0 oil tempered carbon steel wire (show all details for full credit)

7. A stock torsion spring is made of 0.055 is made of 0.055 inch music wire and has N = 6 coils and

straight ends 2 inches long and 180 apart. The OD is 0.654 in.

a. what value of the torque would cause a max. stress equal to the yield of the wire ?

b. compute the angle of rotation corresponding to this torque

c. find the spring ID under the applied torque

d. find the max. allowable torque for infinite life with F.S. = 1.5 provided that the min. torque is always

20% of the max.

Ans./hints

Refer to Shigley for design eqns. for torsion springs:

C =D/d

Sut = A/dm

Sy = 0.78Sut

=Ki 32Fr/d3

Ki = (4C2-C-1)/{4C(C-1)}

T = Fr

h = d4E/{10.8DN}

=n(360)

Di’ = N/(N+n) Di

Goodman:

1/F.S. = a/Se +m/Sut

a. T = 3.535 in-lb

b. 180

c. ID = 0.502 inches

d. Tmax = 1.42 in-lb

8. Consider the U-joint shown below:

where

1 denotes the angle of rotation of axle 1, 2 denotes the angle of rotation of axle 2 and  denotes the

bend angle of the joint. Recall from ME 325 lecture, that the angular velocities between shaft 1 and

shaft 2 denoted by 1 and 2 respectively are related by

𝜔2 = 𝜔1𝑐𝑜𝑠𝛽

1 − 𝑠𝑖𝑛2𝛽𝑐𝑜𝑠2𝛾1

Or after some trigonometric identities are employed, we can write the so-called “transmission ratio”, 

(a.k.a. efficiency of the u-joint) as:

2/1 =𝜂 = 𝑐𝑜𝑠𝛽(1+𝑡𝑎𝑛2𝛾1)

𝑡𝑎𝑛2𝛾1+𝑐𝑜𝑠 2𝛽

whence, we find upon plotting:

and

1,1

2,2

and

Now, consider a situation with the shaft angle is β=30, axle 1 has a speed of n1 = 1000 rpm

(1=104.720 rad/sec) , determine:

a. max. & min. speeds for axle 2, 2,min/2,max

b. max. / min. input shaft 1 angles correspond to max. / min. of part a.

c. for the situation above, if shaft 1 transmits 10 hp, what is the u-joint efficiency ?

d. for the situation above, what is the power (hp) transmitted by shaft 2 ?

1

 2 /

1

9. Determine the mass flow rate through a straight labyrinth seal with the following geometric

characteristics: four fins, n=4,outer diameter od = 2*ri=215.9mm, fin height h=2.286mm, pitch

lpitch=3.175mm, fin taper angle fin=7.5, radial clearance c=0.127mm, fin-tip land width t=0.305mm. The

total pressure drop pt,0 = 3.01105 Pa, the downstream pressure is pn= 1.01105 Pa and the average seal

temperature is T=300 L. R = 287 J/kg-K and viscosity is =1.810-5 Pa-sec.

Use the theory of Zimmerman & Wolf (1998) whereby

𝑚 = 𝑘2𝐶𝑑𝐴𝑝𝑡,0√ 1 − (𝑝𝑛/𝑝𝑡,𝑂)

2

𝑅𝑇[𝑛 − ln⁡(𝑝𝑛/𝑝𝑡,𝑂)]

The carry-over factor is

𝑘2 =

√

𝑛/(𝑛 − 1)

𝑛 𝑛 − 1 −

[ 𝑐/𝑙𝑝𝑖𝑡𝑐ℎ 𝑐

𝑙𝑝𝑖𝑡𝑐ℎ + 0.02

]

The discharge coefficient, Cd depends on the Reynold number as shown below ( as of now Re is

unknown, thus guess and check the assumption later)

The annular seal cross-sectional flow area is 𝐴 = 𝜋(𝑟𝑜 2 − 𝑟𝑖

2);𝑟𝑜 = 𝑟𝑖 + 𝑐; 𝑟𝑖 = 𝐷/2

Proceed to compute m, then use the value found to check Re = mDh/A;Dh= 2(ro-ri), repeat until

converged. Show at least 3 iterations for full credit.

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