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Unformatted text preview: ;" fsTmmi ISL“ @21ka
31' w. G’J‘ﬂm%‘\€yh Ncsymﬂ 05w: g‘)ii—‘W
{33¢}; , 50;] I SEE} ginecw fhegig IW‘n rls, (1(er (if). ESL: ﬁshing 1 'We 14%? “'sz Sui/WE Moi/W q," mks;
(L63: T'=!:’1/LQ [Dr {La “33‘: [Q ‘ﬁieds" g. 915363;: ”5t
Gina/“3%“ '4 MI jﬁ’t we :3 W W4 SW” er W» WMHFQV r \ “EU I. u" U.) %WTLG fS‘wwe " I231 (El/INQ "9’3' BiLSfQ/qDjﬁ’t (3Y1: \Qbﬁxwr, marm. .\N 108 PART III DRY son. b Example 8.4
Given. Figure EBA1. 2
20 kN/m B Find. Stresses on plane B—B.
Solution. Use Fig. 138.42. Fig. EBA2 Locate points with coordinates (40, 0) and (20, 0). Draw circle, using these points to deﬁne diameter. Draw line A’A’ through point (20,0) and parallel to plane on which stress (20, 0) acts.
Intersection of A A with Mohr circle at point (40,0) is the origin of planes. Draw line B "B through 0;: parallel to BB. .Read coordinates of point X where B 'B intersects Mohr circle.
nswer. See Fig 133 43 g' = 25 kN/m2 7 = —8.7 kN/m2 A9999N!‘ on BB { \2‘5 kij2 2' 20 kN/m2 Fig. 8.43 Alternate Solution. Steps I and 2 same as above.
3. Draw line C'C’ through (40, 0) parallel to plane on which stress (40, 0) acts. CC is
vertical. 4. C'C’ intersects Mohr circle only at (40, 0), so this point is Op. Steps 5 and 6 same as
above. Solution Using Eqs. 8.6 and 8.7.
= 40 lcN/rn2 “a H 20kN/m2 6 = 120" 40+20 40—20 2+2 4t) — 20
19 = 2 sin 240° = —10 sin 60" = "8.66 kN/m2 4 do = cos 240° = 30 — 10 cos 60° = 25 kN/m2 (Questions for student. Why is 9 = 120‘“? Would result be different if 8 = 300° ‘2‘) 110 PART III DRY SOIL
Example 8.6
Given. Figure E8.6I. D D
/\ / 10 kN/m2
A/ \ 20 kN/mz
B
Fig. E8.61 Find. Magnitude and direction of the principal stresses.
Solution. Use Fig. E3.6—2. IIIII III
IIIIIIIIII
IIIIIIIIII
IIIIIIIIII
IIIIIIIIII
IIIIIIII II Fig. E8.62 .BI 10 —10 40 50 1. Locate points (40, ”10) and (20, 10). 2. Erect diameter and draw Mohr circle. 3. Draw B’B’ through (40,—10) parallel to BB.
4. Intersection of B’B’ with circle gives OP. 5. Read ‘71 and aa from graph. 6. Line through 01, and 01 gives piano on which 01 acts, etc. (see Fig. 138.63). 0'1  44.1 kN/mz
52' D D
as = 15.9 kN/mz Fig. E8.63 Solution by Equations.
1. First make use of fact that sum of normal stresses is a constant: 6+6 Ea 40+2
1 ’=——"=— =30kN/m2 2 2 2 Ch. 8 Stresses within :1 Soil Mass 111 with either pair of given stresses 2. Use relation (a;  as) =JW = Jiﬁ =14.14k1~1/m2 2
a — a
3. a. = (El—:1?) + (4—33) = 44.14 kN/m2 E
 E
a + a — 0' g
as = (3—5—3) — (‘12—?) = 15.86 kN/m? :
4. Use stress pair in which “a is largest; Le. (40, —10)
_ ' 279 —20
5m 26 — 3:0; — m —— —0.707
26 = 45°
6 = —22;~° ...
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 Spring '16
 steve sun

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