A model boiler must be built for Matimba Power Station, which is one of the major

electricity-generating power stations in South Africa, to operate at a working pressure
of 1 250 kPa. The materials to be used include 10 mm-thick steel plates riveted
together by a double-riveted butt joint. The joint is longitudinal and secured with a
cover plate outside and inside the boiler. 20 mm-diameter rivets are used at a pitch of
60 mm. Both the boiler and rivets are made of the same material.
The following stresses are given with a safety factor of 4:
Tensile: 420 MPa
Shear: 300 MPa
Compressive: 600 MPa
1.1 Calculate the efficiency of the joint.
1.2 Calculate the maximum permissible diameter of the boiler.

2 answers

To calculate the efficiency of the riveted joint and the maximum permissible diameter of the boiler, we will follow these steps:

1.1 Calculate the Efficiency of the Joint

The efficiency of a riveted joint can be calculated using the formula:

\[ \text{Efficiency} (\eta) = \frac{\text{Strength of the joint}}{\text{Strength of the member}} \times 100% \]

Strength of the member (Steel Plate):

The effective area of the plate, which is the net area when considering the rivet holes, must be calculated first.

  1. Net Area of Steel Plate (Anet):

    The diameter of each rivet is given as \(d_r = 20 , \text{mm}\), and the pitch is \(p = 60 , \text{mm}\). The effective area is reduced due to the rivet holes.

    \[ \text{Net Area} (A_{net}) = w \cdot (t - d_r) = w \cdot (t - 20, \text{mm}) \] where \(w\) is the width of the plate, and \(t\) is the thickness of the plate.

    However, we currently lack the width of the plate, so we need additional information or assume a width.

  2. Tensile Strength of the Plate:

    The tensile stress is \(420, \text{MPa}\), and the area can be simplified for calculating efficiency using the gross area approach since we need material properties only:

    \[ \text{Ultimate tensile force} = \sigma_t \cdot A_{gross} \]

    The gross area \( A_{gross} = w \cdot t = w \cdot 10 , \text{mm}\).

  3. Strength of the Rivet:

    The strength of a single rivet in shear:

    \[ \text{Shear strength of rivet} = \tau \cdot A_{rivet} \]

    The area of a rivet can be calculated using:

    \[ A_{rivet} = \frac{\pi}{4} d_r^2 = \frac{\pi}{4} (20 , \text{mm})^2 = \frac{\pi}{4} \cdot 400 \approx 314.16 , \text{mm}^2 \]

    The shear strength of the rivet:

    \[ \text{Shear strength of rivet} = \tau \cdot A_{rivet} = 300 , \text{MPa} \cdot 314.16 , \text{mm}^2 \approx 94248 , \text{N} \]

  4. Now, we set the efficiency:

The ultimate strength of the joint is taken as the lesser of the plate strength and the rivet strength.

Assuming \(N\) rivets are used,

  • Total shear strength = \( N \cdot \text{Shear strength of rivet} \)

So the efficiency is calculated as:

\[ \eta = \frac{ \text{Shear strength of rivets} }{ \text{Tensile Strength of Plate} } \cdot 100% \]

Assuming using 2 rivets,

\[ \text{Strength of joint} = 2 \cdot 94248 \approx 188496N \] And we need an area calculation to finalize the joint efficiency; however widely accepted follows.

1.2 Calculate the Maximum Permissible Diameter of the Boiler

The formula for hoop stress \(\sigma_h\) for a thin-walled cylinder is given by:

\[ \sigma_h = \frac{P \cdot D}{2 \cdot t} \]

Where:

  • \(D\) = Diameter of the boiler
  • \(P\) = Internal pressure = 1250 kPa = \(1250 \times 10^3 , \text{Pa}\)
  • \(t\) = Thickness of the steel plate = 10 mm = 0.01 m

To find the maximum permissible diameter, set \(\sigma_h = 420 \text{MPa}\):

\[ 420 \times 10^6 = \frac{(1250 \times 10^3) \cdot D}{2 \cdot 0.01} \]

Rearranging gives:

\[ D = \frac{2 \cdot 0.01 \cdot (420 \times 10^6)}{1250 \times 10^3} \]

Now simplifying:

\[ D \approx \frac{2 \cdot (420 \times 10^6) \cdot 0.01}{1250 \times 10^3} = \frac{8400000}{1250} \approx 6720 , \text{mm} \approx 6.72 , \text{m} \]

Conclusion:

  1. The efficiency of the joint would need the final values considering the number of rivets.
  2. The maximum permissible diameter of the boiler is approximately 6.72 m.

Would you like further elaboration on riveted joint efficiency calculation or any other questions?

Data:
t = 10 mm; n = 2; p = 60 mm; o+ = -
300
- МРа = 105 МРа ; т = — МРа = 75 МРа
0c = 600
- MPa = 150 MPa d = 20mm
4
Solution:
1.1 The efficiency of the joint:
F+ = (p - d)to+-p = (60 - 20)(10) (105)
= 42 kN
Fs=Fa7TR=(20) (75) (2)
= 47,124 kN
Fc = dtnoc-PR = (20) (10) (2) (150)
PC-18
= 60 kN
Fo = ptot-p = (60) (10) (105) = 63 kN
Ft 42
n = - × 1 0 0 % = 4
Fo
63 × 100%
= 66,667%

1.2
The maximum permissible diameter of the boiler:
pD
0c = 2tnL
iD = 2tochL
2(0,01)(105 × 106)Y (66,667 × 10-3)= 1,12 mr
1 250 × 103