Holooly Plus Logo
Holooly Plus Logo

Question 19.6: Which of the following two methods of manufacturing a produc...

Which of the following two methods of manufacturing a product 150 mm in length would give the best overall tolerance?

1. An assembly of 10 components, each of length 15.000 mm with tolerance limits of ± 0.050 mm.

2. A single component of 150 mm also with tolerance limits of ± 0.050 mm.

The blue check mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

1.               \sigma_{1}=0.05 / 3

 

\sigma_{z}^{2}=10 \sigma_{1}^{2}

 

\sigma_{z}=0.0527

Tolerance limits for overall length ± 3σ = ± 0.158 mm.

\mu_{z}=10 \mu_{1}=150.000 mm

 

z=150.000 \pm 0.158 mm

 

2.             z=150.000 \pm 0.050 mm

As this method gives smaller tolerance limits, this is the most suitable way of achieving the tighter tolerance.

Related Answered Questions

Question: 19.1

Find the shaft and hole dimensions for a loose running fit with a 35-mm-diameter basic size. ...

Verified Answer:

From Table 19.1, the ISO symbol is 35H11/c11 (the ...
Question: 19.2

Find the shaft and hole dimensions for a medium drive fit using a basic hole size of 60 mm. ...

Verified Answer:

From Table 19.1, the ISO symbol is 60H7/s6. Table ...
Question: 19.9

The torque capacity, T, for a multiple disc clutch can be calculated using the relationship T =2/3 µ FaN ( r³ o – r³/ r² o – r²) where µ is the coefficient of friction, Fa is the axial clamping force, N is the number of disc faces, ro is the outer radius of clutch ring, and ri is the inner radius ...

Verified Answer:

T_{a v}=\frac{2}{3} \times 0.3 \times 4000 ...
Question: 19.8

In the design work for a model single cylinder reciprocating engine, the work done during the expansion stroke is to be represented as: W = PπD²C ( 1 – Q^n-1) / (4(n – 1)) where Q = C/(C + 2R) and C = M – R – L – H The quantities are defined as follows: W is the work done during the expansion ...

Verified Answer:

The sure-fit “extreme” limit of the work done is g...
Question: 19.7

The stiffness of a helical compression spring (see Figure 19.12) can be calculated from k = d^4G/8D³n where d is the wire diameter, G is the modulus of rigidity, D is the coil diameter, n is the number of coils, and k is the spring stiffness. The mean and standard deviation for each variable are ...

Verified Answer:

Substituting values into the equation for k to giv...
Question: 19.4

A product is made by aligning four components (see Figure 19.11) whose lengths are x1, x2, x3, and x4. The overall length is denoted by z. The tolerance limits of the lengths are known to be 8.000 ± 0.030, 10.000 ± 0.040, 15.000 ± 0.050, and 7.000 ± 0.030 mm, respectively. If the lengths of the ...

Verified Answer:

Assuming ±3σ natural tolerance limits: \sig...
Question: 19.10

For a standard series “A” Belleville spring, the spring rate or stiffness can be determined by the equation k = 4E/(1 – µ²) × t³/K1D²e where E is Young’s modulus of the washer material, t is the washer thickness, m is Poisson’s ration,K1 is a dimensionless constant, and De is the external diameter ...

Verified Answer:

k=\frac{4 E}{\left(1-\mu^{2}\right)} \times...
Question: 19.5

Suppose the statistical tolerance bandwidth for the overall length of the previous example (±0.077 mm) is too large for the particular application. An overall statistical tolerance of ±0.065 mm is required. How can this be achieved? ...

Verified Answer:

This can be achieved in several different ways. On...
Question: 19.3

What interference fit is required in order to transmit 18 kN m torque between a 150-mmdiameter steel shaft and a 300-mm-outside-diameter cast iron hub that is 250 mm long? (Take Young’s modulus for the steel and cast iron to be 200 and 100GPa respectively,and assume the coefficient of friction is ...

Verified Answer:

The torque is given by T=2 f p \pi b^{2} L[...