 Resistanct of the Wall. — Considering the wall as a solid mass, the effect of its weight to resist an overturning thrust will be directly as the horizontal distance EH from its front edge to a vertical line drawn through G, the centre of gravity of...  Elementary Applied Mechanics - Page 58by Thomas Alexander - 1916 - 512 pagesFull view - About this book
 | John Bullock - Architecture - 1853 - 192 pages
...resist an overturning thrust will be directly as the horizontal distance BH from its front edge to a vertical line drawn through G-, the centre of gravity of the wall, fig. 13 ; or, calling the resistance B, and the weight of the wall w, then R = w> XE H. EH will be... | |
 | Ed Dobson - Building - 1854 - 182 pages
...resist an overturning thrust will be directly as the horizontal distance EH from its front edge to a vertical line drawn through G, the centre of gravity of the wall, fig. 18; or calling the resistance R, and the weight of the wall w, then E=wx E H. EH will be directly... | |
 | John Bullock - Architecture - 1855 - 508 pages
...resist an overturning thrust will be directly as the horizontal distance EH from its front edge to a vertical line drawn through G, the centre of gravity of the wall, fig. 13 ; or, calling the resistance R, and the weight of the wall w, then R = w XE H. EH will be directly... | |
 | Oliver Byrne - 1867 - 380 pages
...the pressure is sufficient to overturn the embankment it will turn upon A, as a centre ? Let FOC be the vertical line drawn through G, the centre of gravity of the embankment. Draw PL perpendicular to FC, intersecting FC in O. Make O n= the Ibs. pressure in the embankment,... | |
 | Edward Spon - Engineering - 1870 - 402 pages
...the pressure in sufficient to overturn the embankment it will turn upon A, as a centre. Let FOG be the vertical line drawn through G, the centre of gravity of the embankment. Draw PL perjiendicular to FC, intersecting F С in О. Make О я = the Ibs. pressure in... | |
 | Thomas Alexander (civil engineer.) - 1880 - 196 pages
...BA = 3 . Q tends to overturn the wall with a moment, M = Q x leverage about B, D sin . foot _i bs . the centre of gravity of the wall, cut the joint at s, then for equilibrium the moment, the weight of wall x leverage KS, must be greater than the overturning moment, M. It is generally sufficient... | |
 | Edward Dobson - Building - 1890 - 248 pages
...resist an overturning thrust will be directly as the horizontal distance EH from its front edge to a vertical line drawn through G, the centre of gravity of the wall, fig. 13; or calling the resistance K, and the weight of the wall w, then R=«J x E H. EH will be directly... | |
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