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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Side 4
... taking moments about the other end ; or it may be found at once since we know P + Q by equation ( 1 ) . An uniform load , such as w lbs . per running foot spread over a portion of span , is to be treated as one force equal to the amount ...
... taking moments about the other end ; or it may be found at once since we know P + Q by equation ( 1 ) . An uniform load , such as w lbs . per running foot spread over a portion of span , is to be treated as one force equal to the amount ...
Side 9
... Taking moments about S , the moment of Σ ( W ) acting at a will equal the sum of the moments of W1 , W ... acting at a ,, a ..... That is Σ ( W ) × ā = Σ ( Wa ) ; ā = Σ ( Wa ) Σ ( W ) gives position of G measured from S the left end of ...
... Taking moments about S , the moment of Σ ( W ) acting at a will equal the sum of the moments of W1 , W ... acting at a ,, a ..... That is Σ ( W ) × ā = Σ ( Wa ) ; ā = Σ ( Wa ) Σ ( W ) gives position of G measured from S the left end of ...
Side 30
... Taking the centre of span as origin , the abscissa of P is c ; and , if x be the abscissa of the point about which moments are taken , then ( c - x ) is the leverage of P , and P tends to break the beam at the point by bending the left ...
... Taking the centre of span as origin , the abscissa of P is c ; and , if x be the abscissa of the point about which moments are taken , then ( c - x ) is the leverage of P , and P tends to break the beam at the point by bending the left ...
Side 41
... Taking the origin at the centre , as Rankine does , M. equals the bending moment at the centre ; for loading symmetrical about the centre M = Maximum Bending Moment . Examples . 14. A beam 24 feet span is loaded with 20 , 30 , and 40 ...
... Taking the origin at the centre , as Rankine does , M. equals the bending moment at the centre ; for loading symmetrical about the centre M = Maximum Bending Moment . Examples . 14. A beam 24 feet span is loaded with 20 , 30 , and 40 ...
Side 42
... taking the centre as origin , M2 = 256 ft . - tons , maxi- mum , and is at the wheel transmitting W , 11 tons . 3 = Thus for the position of the locomotive given in No. 12 , that is with its fore wheel 6 feet from the left end , the ...
... taking the centre as origin , M2 = 256 ft . - tons , maxi- mum , and is at the wheel transmitting W , 11 tons . 3 = Thus for the position of the locomotive given in No. 12 , that is with its fore wheel 6 feet from the left end , the ...
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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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