The Mechanics of Construction, EtcDughton, Bull & Company, 1861 |
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Page xi
... CIRCULAR SECTIONS . Cylindrical beams . Hollow cylindrical beams Summary of formulæ for cylindrical beams Examples . 63 63 64 64 The section a triangle Section 4 . BEAMS OF DIFFERENT KINDS . The section a parallelogram The section a ...
... CIRCULAR SECTIONS . Cylindrical beams . Hollow cylindrical beams Summary of formulæ for cylindrical beams Examples . 63 63 64 64 The section a triangle Section 4 . BEAMS OF DIFFERENT KINDS . The section a parallelogram The section a ...
Page xiii
... CIRCULAR ARCHES WITHOUT SURCHARGE . PAGE 119 • 120 121 121 122 122 123 123 124 124 128 130 131 134 134 134 • 135 137 137 138 139 Conditions of equilibrium of a circular arch Thickness of pier necessary to support a semicircular arch ...
... CIRCULAR ARCHES WITHOUT SURCHARGE . PAGE 119 • 120 121 121 122 122 123 123 124 124 128 130 131 134 134 134 • 135 137 137 138 139 Conditions of equilibrium of a circular arch Thickness of pier necessary to support a semicircular arch ...
Page xiv
Stephen Fenwick. Section 3 . CIRCULAR ARCHES WITH SURCHARGE . Conditions of equilibrium of a circular arch with a horizontal surcharge . Strength of pier Examples • Section 4 . THE STABILITY OF THE ARCH IN GENERAL . CURVE OF PRESSURE ...
Stephen Fenwick. Section 3 . CIRCULAR ARCHES WITH SURCHARGE . Conditions of equilibrium of a circular arch with a horizontal surcharge . Strength of pier Examples • Section 4 . THE STABILITY OF THE ARCH IN GENERAL . CURVE OF PRESSURE ...
Page 19
... circular section of the cylinder ACB . Then it is clear that the centre of gravity of the arc of the semicircle ACB I will be at the same distance from AB as the centre of gravity of the half surface of the cylinder . Refer the elements ...
... circular section of the cylinder ACB . Then it is clear that the centre of gravity of the arc of the semicircle ACB I will be at the same distance from AB as the centre of gravity of the half surface of the cylinder . Refer the elements ...
Page 21
... circular areas of its base is π ( r + t ) 2 — π p2 - = πt ( 2r + t ) , if S be the coefficient of resistance , the total resistance of the base will be πt S ( 2r + t ) • • ( 2 ) . Hence by ( 1 ) and ( 2 ) περ = πt S ( 2r + t ) , and ...
... circular areas of its base is π ( r + t ) 2 — π p2 - = πt ( 2r + t ) , if S be the coefficient of resistance , the total resistance of the base will be πt S ( 2r + t ) • • ( 2 ) . Hence by ( 1 ) and ( 2 ) περ = πt S ( 2r + t ) , and ...
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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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