Elementary Applied Mechanics: Being the Simpler and More Practical Cases of Stress and Strain Wrought Out Individually from First Principles by Means of Elementary Mathematics, Volum 2Macmillan and Company, 1883 |
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Resultat 1-5 av 55
Side xi
... length , Cantilever do . do . , Examples , 40-46 , - Ordinates of parabola and of triangle added and subtracted , Beam uniformly loaded on its two segments with loads of different intensity , Beam do . do . Beam uniformly loaded and ...
... length , Cantilever do . do . , Examples , 40-46 , - Ordinates of parabola and of triangle added and subtracted , Beam uniformly loaded on its two segments with loads of different intensity , Beam do . do . Beam uniformly loaded and ...
Side xii
... length , 184 184 185 do . , 185 185 Beam do . do . , 186 Beam loaded uniformly on two segments with different intensities of load , 186 Shearing Forces and Shearing Force Diagrams for Moving Loads : - Range , Definition of , 187 Beam ...
... length , 184 184 185 do . , 185 185 Beam do . do . , 186 Beam loaded uniformly on two segments with different intensities of load , 186 Shearing Forces and Shearing Force Diagrams for Moving Loads : - Range , Definition of , 187 Beam ...
Side 2
... length 2c , depth h , and breadth b , and OX is any line chosen as axis . W , is a force in the plane of the paper , ] W2 each w O B B ' W1 P Q Fig.1 . replacing a stress spread uniformly over the breadth of the beam , as shown on the ...
... length 2c , depth h , and breadth b , and OX is any line chosen as axis . W , is a force in the plane of the paper , ] W2 each w O B B ' W1 P Q Fig.1 . replacing a stress spread uniformly over the breadth of the beam , as shown on the ...
Side 3
... length of the beam . Since the forces are all parallel and in one plane , there are two conditions of equilibrium : -- I. The algebraic sum of the forces is zero . II . The algebraic sum of the moments of the forces about any point is ...
... length of the beam . Since the forces are all parallel and in one plane , there are two conditions of equilibrium : -- I. The algebraic sum of the forces is zero . II . The algebraic sum of the moments of the forces about any point is ...
Side 4
... length into four equal parts . Find the supporting forces . = W1 = W 、 x1 = 20 ; W 30 ; W , 40 . 2 6 ; x2 3 = 12 ; X3 18 ; 2c 24 . = = Σ ( Wx ) Q = 2c 20 × 6+ 30 × 12 + 40 x 18 24 = 50 tons . P = Σ ( W ) - Q = ( 20 + 30 + 40 ) - 50 40 ...
... length into four equal parts . Find the supporting forces . = W1 = W 、 x1 = 20 ; W 30 ; W , 40 . 2 6 ; x2 3 = 12 ; X3 18 ; 2c 24 . = = Σ ( Wx ) Q = 2c 20 × 6+ 30 × 12 + 40 x 18 24 = 50 tons . P = Σ ( W ) - Q = ( 20 + 30 + 40 ) - 50 40 ...
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Elementary applied mechanics, by T. Alexander (and A.W. Thomson). Thomas Alexander (civil engineer.) Uten tilgangsbegrensning - 1883 |
Vanlige uttrykk og setninger
abscissa advancing load apex beam bending moment diagram bh³ calculate the bending cantilever centre of gravity centre of span chord College common interval construct cross section Crown 8vo curved edge passes dead load distance draw a vertical drawn Edition end of span equal example fcap feet span field Find the maximum fixed end fixed loads ft.-tons gives the maximum graphical solution greater greatest hence horizontal inch intersection left end leverage locus M₁ maxima bending moments maximum bending maximum bending moment modulus moment of inertia negative neutral axis ordinate parabola parabolic segment parallel plane polygon portion position rectangle right end rolling load scale for verticals shearing force shearing stress shown in fig side slope stress Suppose tangent total load transit triangle U+2R uniform load uniformly W₁ W₂ weights wheel whole load whole span
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