The boiler by Christie Stephen 1849-

Author:Christie, Stephen, 1849-
Language: eng
Format: epub
Tags: Steam-boilers
Publisher: Chicago, Ill., Christie Publishing Co.
Published: 1908-03-25T05:00:00+00:00

Causes for failure at joint.

1st. Resistance to shearing three rivets.

2nd. Resistance to tearing between three rivets.

3rd. Resistance to crushing in front of three rivets.

Assuming a boiler of dimensions and data as follows: Legend:

T =thickness of plate = 3^ = . 375 TS =tensile strength = 55000

d -diameter of rivet = 13/16 = . 8125

A =area of rivet hole = 13 a6 = . 5185

P =pitch of rivet = 3 3^^ = 3. 2500 SR = shearing resistance of rivets =38000 CS = crushing strength of rivet and plate =95000

D = diameter of boiler = 60"

F = factor of safety = 5

First. Resistance to shearing of three rivets.

Rule to find strength of rivets in single shear: Multiply area of rivet hole by number of rivets, and multiply this sum by the shearing resistance of rivet material.

Formula:

Ax No. of rivets XvSR= strength of rivets in single shear

Example:

.5185 =area of rivet hole 3 = number of rivets

1.5555

38000 = shearing resistance of rivets

124440000 46665

59109.0000 59,109 lbs. = strength of three rivets in single rhear

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Second. Resistance to tearing of plate between three rivets.

Rule to find strength of net section of plate: From pitch of rivets subtract diameter of rivet hole and multiply by thickness of plate and multiply this sum by the tensile strength of plate.

Formula:

(P—d) XTxTS=strength of net section of plate

Example:

3. 2500 =pitch of rivet .8125 = diameter of rivet hole

2.4375

.375 = thickness of plate

121875 170625 73125

.9140625

55000 = tensile strength

45703125000 45703125

50273.4375000 50,273 = strength of net section of plate

Third: Resistance to crushing in front of plate in front of three rivets.

Formula:

dX3xTxCS= resistance to crushing in front of three rivets

Example :

.8125 = diameter of rivet 3 = three rivets

2.4375

.375 = thickness of plate

121875 170625 73125

.9140625

95000= crushing strength of rivet

and plate

4570 3125000 82265 625

86835.9375000 86,835 lbs. = resistance to crushing of material

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Rule to find strength of solid plate: Multiply pitch of rivets by thickness of plate and this sum by tensile strength of material.

Formula: P X T X TS = strength of solid plate

Example:

3.2500=pitch

.375 = thickness of solid plate

162500 227500 97500

1.2187500

55000 = tensile strength

6093 7500000 60937 500000

67031.2500000 67,031 lbs. = strength of solid plate

Rule to find efficiency of this joint: Divide net section of plate by strength of solid plate.

Example:

50,273 =net section of plate 67,031 = strength of solid plate

67031)50273.000 (. 749 =efficiency 46921 7

3351 30 2681 24

670 060 603 279

66 781

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LAP JOINTS.

153

Rule to find safe working pressure from these calculations: Multiply tensile strength of plate by efficiency of joint and multiply this sum by twice thickness of plate; divide this product by diameter of boiler in inches multiplied by factor of safety.

Example:

55000 = tensile strength of plate . 749 = percentage of joint

495 000 2200 00 38500 0

41195.0^^

.7500 = twice thickness of plate

diam. of boiler = factor of safety =

=60" 2059 7500 = 5 28836 5

300)30896.;2^^^(102.9 lbs. working pressure 300

896 600

2962 2700

262

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CHAPTER VII.

BUTT JOINT DOUBLE STRAPPED AND DOUBLE RIVETED.

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