Question

For the stress state shown, use Mohr's Circle to determine the following:

 (a) Principal stresses σ1 and σ2, the associated shear stress τx'y', and the element orientation θp. Clearly indicate magnitude and direction.

 (b) Maximum in-plane shear stress τmax,in plane , the associated normal stresses σx' and σy', and the element orientation θs. Clearly indicate magnitude and direction.

 (c) Absolute maximum shear stress τmax,absolute · 

Show all work using Mohr's Circle. If your drawings do not indicate that you used Mohr's Circle appropriately, you will receive zero points.

Answer 1

Oy = -25ksi Iny=15ksi ܣ 8y = 18k si Maximum, min. Principle stress e 0,,2. [(0049) + / 05-03*+ 42x] 0112 [118-25) + V (18+251 27 152 ] are 5 28.715 KSI 29.715 ksi Radius of noho Cicle = 0,002 R=227 +295. 26.215 ksi Center at: (-29.7184 26:215,0) = (-3.5,0)

o absolute Imore 3 absolute Zimpleme Mohr circle a (-25, 15) (18,15 > 1 Spor o Or (-29.7,0) Ć 227, ) (-3.5,0) Q tan 20p=2 tony 2x z oy-by 18425 08=147.45", 107.45 01= 22.71 , 09=+29.71 K31 (b timaalinpiame = R = 26.2156879 (at point P,) P= 626.215, -3,5) @ 62.45º 10 Q = (1626. 215, -3.5)@152.450 os In this case 26215k88

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Answer 2

125 25ksi Now Cordinates of mohr circle for give Strosses Az (18,15) B=(-25,-15) -18ksi 118651 1 f 15ksi 25 ksi Now form Mohr م کے Iman AC18,15) zos is 20p NP, ((P2,0) (8,0) C-25,-15) NB 55.93 ksi 15 to 29 2 a) RAJ[18=(-233]+(15-1-153) z 52.43 TOP,- -3.5 + 52.43 - 48.93 ksi Jpnir Principl stresses Fz=P2.-3.5-52.435 - Tx'yl = 0ksi (associated with plane of P, RP₂) Op aletan I tant ( 215 tis) 34.9 17.45° (clockwise) 3.5+18) 16). Tmax R = 52.43ksi (Planne) Normal stress associated with et = -3.5ksi as sostel 45° - Op = 450-17.45º = 27.55° Canticlaekwise) Max {179, 10), 5:)] [148 2:07,1-5597-01, 148*43+53:)] (6) Tmax, a absolute = - Max 52.43 ksi Ans.

from fig2 we found that on rotating the element of fyl by Op we get principle planes, now see fig3 to get P, &k (Principle planes) P, minor principle plane Pes mejor princapleplane 3.51 бе. کم C(-3.5,0) Piccolo batt t r r fy3 OP. = 7-3.5=26.2154-3.5 6p, -22.7154 Kei tensile) OP_=+62+3.526.2154+3.5) 8p=- 29.7154 ksi (compressive av? Tik-- v PL Top x! x 6p=22.7154 element2 op2=29.7154 ksi Major principle stress of. =-29.7154 ki (compressive) Minor principle stress op = 22.7154 Kit (tensile) Shear stress , Teriyl= Op 17.4512° clockwise all dificted on elements Answer Partly

Answer 3

125 25ksi Now Cordinates of mohr circle for give Strosses Az (18,15) B=(-25,-15) -18ksi 118651 1 f 15ksi 25 ksi Now form Mohr م کے Iman AC18,15) zos is 20p NP, ((P2,0) (8,0) C-25,-15) NB 55.93 ksi 15 to 29 2 a) RAJ[18=(-233]+(15-1-153) z 52.43 TOP,- -3.5 + 52.43 - 48.93 ksi Jpnir Principl stresses Fz=P2.-3.5-52.435 - Tx'yl = 0ksi (associated with plane of P, RP₂) Op aletan I tant ( 215 tis) 34.9 17.45° (clockwise) 3.5+18) 16). Tmax R = 52.43ksi (Planne) Normal stress associated with et = -3.5ksi as sostel 45° - Op = 450-17.45º = 27.55° Canticlaekwise) Max {179, 10), 5:)] [148 2:07,1-5597-01, 148*43+53:)] (6) Tmax, a absolute = - Max 52.43 ksi Ans.

from fig2 we found that on rotating the element of fyl by Op we get principle planes, now see fig3 to get P, &k (Principle planes) P, minor principle plane Pes mejor princapleplane 3.51 бе. کم C(-3.5,0) Piccolo batt t r r fy3 OP. = 7-3.5=26.2154-3.5 6p, -22.7154 Kei tensile) OP_=+62+3.526.2154+3.5) 8p=- 29.7154 ksi (compressive av? Tik-- v PL Top x! x 6p=22.7154 element2 op2=29.7154 ksi Major principle stress of. =-29.7154 ki (compressive) Minor principle stress op = 22.7154 Kit (tensile) Shear stress , Teriyl= Op 17.4512° clockwise all dificted on elements Answer Partly

Answer 4

125 25ksi Now Cordinates of mohr circle for give Strosses Az (18,15) B=(-25,-15) -18ksi 118651 1 f 15ksi 25 ksi Now form Mohr م کے Iman AC18,15) zos is 20p NP, ((P2,0) (8,0) C-25,-15) NB 55.93 ksi 15 to 29 2 a) RAJ[18=(-233]+(15-1-153) z 52.43 TOP,- -3.5 + 52.43 - 48.93 ksi Jpnir Principl stresses Fz=P2.-3.5-52.435 - Tx'yl = 0ksi (associated with plane of P, RP₂) Op aletan I tant ( 215 tis) 34.9 17.45° (clockwise) 3.5+18) 16). Tmax R = 52.43ksi (Planne) Normal stress associated with et = -3.5ksi as sostel 45° - Op = 450-17.45º = 27.55° Canticlaekwise) Max {179, 10), 5:)] [148 2:07,1-5597-01, 148*43+53:)] (6) Tmax, a absolute = - Max 52.43 ksi Ans.

from fig2 we found that on rotating the element of fyl by Op we get principle planes, now see fig3 to get P, &k (Principle planes) P, minor principle plane Pes mejor princapleplane 3.51 бе. کم C(-3.5,0) Piccolo batt t r r fy3 OP. = 7-3.5=26.2154-3.5 6p, -22.7154 Kei tensile) OP_=+62+3.526.2154+3.5) 8p=- 29.7154 ksi (compressive av? Tik-- v PL Top x! x 6p=22.7154 element2 op2=29.7154 ksi Major principle stress of. =-29.7154 ki (compressive) Minor principle stress op = 22.7154 Kit (tensile) Shear stress , Teriyl= Op 17.4512° clockwise all dificted on elements Answer Partly

720 +34.9025 =90 28, 355-0979 anticlockwise from Ti (-3.5, Tman) ; Tz (-3.5,-€ mese) obtained by AC les (-3. Stoman? from fig 4 Tmane=r=26.2154 ki Ox r 2015 1128 6 63.5,0) 205 &note Tmane Clockwise positie Casperour r fig4 sign convention Tz Y'(-3.5 Imen) from fg 4 203+20p = 90° Os=27.5488 xanis Pants T&T ake man shear planes rotating element of fg @ by Os= 27.5488 anticlockwise Tman=&=21.2154 ksi ya 3.5ksi -> x! Tz 3.5ksi Os=27-5488 21.2154 ksi 21,21542 3.5ksi. 3.5ksi element

Tman inplane = 21.2154 ksi = -3.5ksi (compressive) Os=27.5488° Gy'= -3.5ksi (compressive) anticlockwise all dipicted on element3 directions Answer Partly (C) Timex absolute Let op, oh, oo are principle stresses ind, y, 2 Op = 22.7154 km I already found 662 = -29.7154 Ksi stressin direction) Tmane absolute = mane of opo-6/2) =mase of 122.7154+29.454 22-7154-0 29-7154-0 2 = man of (26.2154, 11.3577, 14.8577) Kit tman, absolute = 26.2154 ksi Answer party %=0 Chamond opop ) (19.764),1987 1 sen 6h6p, 2 ) 2 2 / 2

Answer 5

Answer 6

Feel free to ask any doubt in the comment section.

125 25ksi Now Cordinates of mohr circle for give Strosses Az (18,15) B=(-25,-15) -18ksi 118651 1 f 15ksi 25 ksi Now form Mohr م کے Iman AC18,15) zos is 20p NP, ((P2,0) (8,0) C-25,-15) NB 55.93 ksi 15 to 29 2 a) RAJ[18=(-233]+(15-1-153) z 52.43 TOP,- -3.5 + 52.43 - 48.93 ksi Jpnir Principl stresses Fz=P2.-3.5-52.435 - Tx'yl = 0ksi (associated with plane of P, RP₂) Op aletan I tant ( 215 tis) 34.9 17.45° (clockwise) 3.5+18) 16). Tmax R = 52.43ksi (Planne) Normal stress associated with et = -3.5ksi as sostel 45° - Op = 450-17.45º = 27.55° Canticlaekwise) Max {179, 10), 5:)] [148 2:07,1-5597-01, 148*43+53:)] (6) Tmax, a absolute = - Max 52.43 ksi Ans.