The Mechanics of Construction, EtcDughton, Bull & Company, 1861 |
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Page xi
... triangle Section 4 . BEAMS OF DIFFERENT KINDS . The section a parallelogram The section a trapezoid Section with flanges ( Hodgkinson beam ) Resistance to extension and compression Examples CHAPTER VII . SOLIDS OF EQUAL RESISTANCE ...
... triangle Section 4 . BEAMS OF DIFFERENT KINDS . The section a parallelogram The section a trapezoid Section with flanges ( Hodgkinson beam ) Resistance to extension and compression Examples CHAPTER VII . SOLIDS OF EQUAL RESISTANCE ...
Page 29
... triangles GLO and Ln N , we have Nn X · ( 1 ) . S But before flexure , all the fibres comprehended between EF and HK , had for common length GL = s ; whence it follows that the fibre in the direction MN has been elongated by Nn , and ...
... triangles GLO and Ln N , we have Nn X · ( 1 ) . S But before flexure , all the fibres comprehended between EF and HK , had for common length GL = s ; whence it follows that the fibre in the direction MN has been elongated by Nn , and ...
Page 37
... elongation of the fibre passing through the centre of gravity C , x the distance of the element at M from C , and h the distance of the neutral point G from C. Then , by similar triangles , Nn Ln x + RESISTANCE TO FLEXURE . 37.
... elongation of the fibre passing through the centre of gravity C , x the distance of the element at M from C , and h the distance of the neutral point G from C. Then , by similar triangles , Nn Ln x + RESISTANCE TO FLEXURE . 37.
Page 38
Stephen Fenwick. Then , by similar triangles , Nn Ln x + h Dd = Ld = ; h the lines GL , CD , MN being parallel to each other , and ef parallel to HK . Hence x + h Nn Dd . Dividing the left - hand member of this equation by Mn , and the ...
Stephen Fenwick. Then , by similar triangles , Nn Ln x + h Dd = Ld = ; h the lines GL , CD , MN being parallel to each other , and ef parallel to HK . Hence x + h Nn Dd . Dividing the left - hand member of this equation by Mn , and the ...
Page 52
... triangle , of which the equal sides AB and BC are denoted by d . Draw in the triangle ABC the lines DE , FG , parallel to BC ; then evidently AD = DE . Wherefore if we put A D or DE = y , then y'dy or y x ydy , will be the moment of the ...
... triangle , of which the equal sides AB and BC are denoted by d . Draw in the triangle ABC the lines DE , FG , parallel to BC ; then evidently AD = DE . Wherefore if we put A D or DE = y , then y'dy or y x ydy , will be the moment of the ...
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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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