REP17802

• Permit No: C1981016
• IBM Index Class Name: Report
• Doc Date: 10/16/1980
• From: WESTERN ENGINEERS INC
• To: HIGHLAND ENGINEERING
• Permit Index Doc Type: STABILITY REPORT
• Media Type: D
• Archive: No

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Y `j ~ r c

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~ Ce

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= _ n,~ ~ m
`n,
ll 53° _ 5.52
~``\J`
'` O~ Toe circles
-•-•- Vid point circles
---- 31ope circles
00° 60° .0° 00° 50° a0° 30" 2U' tlT u-
B
F'i~. (i.21i. Stability number and types of
slope failures (or ~ = 0. [After Terzaghi
and Pack (1948).
h:curn plr ti.l. The simple slope shown in Fig. 8.23 has a height of 35 ft and a
~!ope I venical to I horizontal. The soil progenies are as follows: y = 12S pcf,
q = IS deg, c = 700 ps(. By means of the stabili[y chart, determine the factor of
,afrq~ with respect to failure.
c„lutlune For a slope of I venical to 1 horizomal we have
tan 13 = 1.0 ~ = 45°
The stability number for ~ = IS deg and ~ = 45 deg is 11.6 (Fig. 8.24). Thus at
:ailure
N.=YH=1I.6
c
1=ar~she given conditions we have
"' 125 X 3S
° N, = 700 = 6.25
.ec. x.s snore a~sr.~~sr~ 2d7
which is much smaller th;m N, required for failure. A sitbility number of I1.6 can
be reached if the height or density is increased or if the cohc~ion is reduced. Failure
would occur if
Y = N,c _ I1.6 X 700 = 232 pcf
N 35
or N=N,c_ 11.6X700=65 (t
y 125
The factor o(safety is then the density (or height) th;tt the soil can support, divided
by the given density (or height):
F' = I25 35 = 1.86
The cohesion required to make N, equal 11.6 if y and N are ai given is
c _ vH _ 1211 635 = 378 psf
The factor of safety may also be the ratio of the given cohesion to that required for
equilibrium, or
~F, = 3~8 = 1.86
It should be noted that this safety factor is for cohesion only. The value o(p used
to enter the chart is IS deg; hence the s;tfety factor with respect to m is I.U. There-
fore the value of 1.55 is not the (actor of safety defined by Eq. (SAO), which con-
siders the entire shear strength.
The safety factor that applies equally to both r ;tnd ~ may be expressed as
F _c+otan~
T
C ran
To calculate this safety factor, we first estimate F, ro be 1.60. Then
tan ¢ _ tan IS° = 0 168
F, 1.60
This corresponds to a ¢ of 9.5 deg. N'e enter the chart (Fig. 8.24) with ~ of 9.5 deg
and find
N. = 9.0 c = 12S9X0 35 = 488 psf
F, = 488 = 1.43
Thus we can have factors of safety of 1.60 (or 0 and 1.43 for c instead of I.00 for
~ and 1.85 for c. If we want the same safety (actor for c and ~, we must revise our
first estimate of 1.60. Subsequent trials give a value of 1.49 for F, that holds for ~~
both c and ¢.