The Mechanics of Construction, EtcDughton, Bull & Company, 1861 |
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Page 15
... diameter . 2. Determine the diameter of the rings of an iron chain capable of raising a weight of 90 cwt . , the coefficient of absolute strength of the iron being taken at 23 tons . Ans . Diameter of each ring 35 of an inch . 3. Find ...
... diameter . 2. Determine the diameter of the rings of an iron chain capable of raising a weight of 90 cwt . , the coefficient of absolute strength of the iron being taken at 23 tons . Ans . Diameter of each ring 35 of an inch . 3. Find ...
Page 16
... Diameter = 4.38 inches . 4. Determine the dimensions of a square column of cast iron in order that it may support a load of 40,568 lbs . in the direction of its axis , the length being inconsiderable , and the coefficient of resistance ...
... Diameter = 4.38 inches . 4. Determine the dimensions of a square column of cast iron in order that it may support a load of 40,568 lbs . in the direction of its axis , the length being inconsiderable , and the coefficient of resistance ...
Page 20
... diameter from 12 to 30 " " " " " " from 30 to 48 " " دو 99 from 48 to 70 " " 99 from 70 to 100 " " 99 Values of t ' g - inch . -inch . ğ - inch . -inch . z - inch . 14. According to Fairbairn , the simple rivetings reduce by one - half ...
... diameter from 12 to 30 " " " " " " from 30 to 48 " " دو 99 from 48 to 70 " " 99 from 70 to 100 " " 99 Values of t ' g - inch . -inch . ğ - inch . -inch . z - inch . 14. According to Fairbairn , the simple rivetings reduce by one - half ...
Page 24
... diameter of each bolt . Hence , m being an arbitrary quantity greater than 2 , let us assume 2 TR > mr · n ( 2 ) . Eliminating between ( 1 ) and ( 2 ) , we get n < 42 S mp · ( 3 ) . This formula and the equation ( 1 ) will solve the ...
... diameter of each bolt . Hence , m being an arbitrary quantity greater than 2 , let us assume 2 TR > mr · n ( 2 ) . Eliminating between ( 1 ) and ( 2 ) , we get n < 42 S mp · ( 3 ) . This formula and the equation ( 1 ) will solve the ...
Page 25
... diameter of the piston being half an inch , and the coefficient of resistance of the metal 16,500 lbs . First , to find the pressure p on a unit of surface of the piston , we have Ρ = 560 ( 1 ) 2 π 560 x 16 π and therefore by ( 8 ) ...
... diameter of the piston being half an inch , and the coefficient of resistance of the metal 16,500 lbs . First , to find the pressure p on a unit of surface of the piston , we have Ρ = 560 ( 1 ) 2 π 560 x 16 π and therefore by ( 8 ) ...
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Common terms and phrases
angle axle beam breadth c₁ catenary centre of gravity centre of pressure chain circular arch coefficient of resistance coefficient of safety common catenary compression cosec cross section cubic foot curve of pressure cylinder deflection Denote density determined diameter dimensions Edition equal equation equilibrium exterior forces extrados extremity F. A. Paley Fcap feet fibre find the thickness fixed formula fracture h₁ Hence horizontal line horizontal position horizontal thrust inertia intrados king-post length lever arm load lowest point modulus of elasticity moment of inertia moments neutral axis parabola parallel perpendicular plane preceding prism PROP r+ h r₁ radius rafter rectangular roof section HK segmental arch sin² springing line stability straining force supported supposed surcharge thickness of pier tie-beam torsion trapezoid triangle vertical pressure weight Wherefore
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