Thus W 1 = 15(I)(ISO) = 2250 Ib (10,008
N); W 2 = 0.5(15)(5)(150) = 5625; W 3 =
0.5(15XS)(IOO) = 3750. Then ^W =
11,625 Ib (51,708 N). Also, Jc 1 = 0.5 ft; X 2
= 1 + 0333(5) = 2.67 ft (0.81 m); Jc 3 = 1 +
0.667(5)-433 ft (1.32m).- Compute the resultant
horizontal soil thrust
Compute the resultant horizontal thrust T
Ib of the soil by applying the coefficient
of active earth pressure. Determine the
location of T. Thus BG = 0.333(1S)(IOO)
= 500 Ib/lin ft (7295 N/m); T =
0.5(15)(500) = 3750 Ib (16,680 N); y =
0.333(15) = 5 ft (1.5m). - Compute the maximum
frictional force preventing
sliding
The maximum frictional force Fm =
FICURE 10 !*& ^' wnere M = coefficient of friction.
Or Fm = 0.6(11,625) - 6975 Ib (31,024.8
N). - Determine the factor of safety against sliding
The factor of safety against sliding is FSS = FJT = 6975/3750 = 1.86. - Compute the moment of the overturning and stabilizing forces
Taking moments with respect to C, we find the overturning moment = 3750(5) = 18,750
lb-ft (25,406.3 N-m). Likewise, the stabilizing moment = 2250(0.5) + 5625(2.67) +
3750(4.33) = 32,375 lb-ft (43,868.1 N-m). - Compute the factor of safety against overturning
The factor of safely against overturning is FSO = stabilizing moment, lb-ft (N-m)/over-
turning moment. lb-ft (N-m) = 32,375/18,750 = 1.73.
ANALYSIS OF A SIMPLE SPACE TRUSSIn the space truss shown in Fig. 1 Ia, A lies in the xy plane, B and C lie on the z axis, and
D lies on the x axis. A horizontal load of 4000 Ib (17,792 N) lying in the xy plane is ap-
plied at A. Determine the force induced in each member by applying the method of joints,
and verify the results by taking moments with respect to convenient axes.
Calculation Procedure:- Determine the projected length of members
Let dX9 dy and dz denote the length of a member as projected on the jc, y, and x axes, re-
spectively. Record in Table 3 the projected lengths of each member. Record the remain-
ing values as they are obtained.