{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW 9 - PBQBLEM 5—3 s The components of plane stress at a...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PBQBLEM 5—3 s The components of plane stress at a point p of a material referred to the x'y‘z' coordinate system are 0‘, = -8 MPa, s', = 5 MPa . and 13., = -16 MPa. and the angle a = 20°. Use Mohr's circle to determine the stresses o,“ cry. and I” at point p. SQLUTION: ai+d._—8M?a+6M?a The center of the circle is at : 0' = -_ ‘ 2 2 rc=OMPa = —lMPa The radius of the circle is: . MP r=1/(0';-—Ur)z+(r;y)l= (—8MPa-(—1MPa))2+(—16MPa)’=17.5MPa T( O) The Angi The angle between the horizontal axis and the line PO is: 6’ = 180° -— 66.3” = 7.3.7" .. The coordinates of point P are: c, = -1 MPa + r (cos 73.7”) = -1 MPa + (17.5 MPa) (cos 73.7“) Angl ANs . g. = 3.91 MPa ‘5- The s. = - r (sin 73.7“) = - (17.5 MPa) (sin 73.7”) ms 1. = - 16.8 MPa ANS The normal stress on the y-face of the element (or-coordinate of point 0) is: s, = .1 MPa s r (cos 73.7”) = .1 MPa — (17.5 MPa) (cos 73.7“) ANS me Q. = - 5.91 MPa "- The shov ,. and .. . 3 norrr The cpomonents of plane stress at a point p of a bit during a drilling I. the F operation are 0x = 40 ksi. cry {-30 ksi. and 1,, = 30 ksi. and the . ; 5Q components referred to the xyz coordinate system are c,‘ = 12.5 ksi, ‘ c,‘ = - 2.5 ksi. and rxy' = 45.5 ksi. Use Mohr’s circle to estimate the The angle 6'. SQLUTION: The 40k5i + —30ksi _ . _ a. —=.__‘_——)ss The center of the carcle Is located at: 2 . . Angl 2'( = Oksz .| . . _ . ' , ' - An I The radius of the circle is: r = (40k:i - 5.1m)2 +(301’c51)1 = 46.1ks: Pfi2-5v‘5-5) . g 1 The B for c, is: 8., = tan" [30ksi/(40ksi~5ksi)] = 40.60 ' T fksi) e for a; is: e... = tan" [45.5ksi/(12.5ksi—5ksi)] = 80.6“ , -. ANS The angle between the two orientations is: 29 = 9., — 60' = 40.60 - 80.60 NS 9 = -20° 266 Use Mohr's circle to determine the principal stresses and the maximum in—plane shear stress and show them acting on properly oriented elements. ox = -8 ksi, o, = 6 ksi. and ‘txy = «6 ksi. §QLQTIOE ‘ The center of the circle is at : a +0' ~— ' ‘ 0}: x2 y: Skr12+6krr=_lm, 1:.,=1:W - The radius of the circle is: r = (a; — at): + (W2 = ‘/(- Sksi — arm-)2 + (- emf = 9.22m - We see from Mohr‘s circle that: o1=uc+r=-1ksi+9.22ksi ANS 93 = - 8.22 ksi 62 = cc - r = -1 ksi — 9.22 ksi 6.22ksi 10.22%: 7- ANS 02 = - 10.22 ksi \ / /9.22k5i L ANS lam] = r = 9.22m / C 9.225%? _‘ 20.3 E; \l i], 7; /\ ' PR - 17 Use Mohr's circle to determine the principal stresses and the maximum in-plane shear stress and show them acting on . properly oriented elements. ox = 240 MPa, cry = -120 MPa. and 1,“, = 240 MPa. '_ mm: 0’ + . .. .-' The center of the circle is at : o-c = 4-2-3 = w = 60MPa 11,, = 1,, The radius of the circle is: r = "(0", - aj + (aw)2 =1/(240MPa . some)2 + (240mm)2 a BOOMPa SOMPa . We see from Mohr's circle that: f \ 1 - faith: ‘ ANS 0'1 = ‘7: + r = GOMPa + SOOMPa = 360MHz \ woman I W EOMPa #35 \ \ / .Ns 0'2 = a} - r = 60MPa — 300MPa = —240MPa if 4 \ ANS pm; = r = 30mm: For the state of plane stress or, = 8 ksi, c:y = 6 ksi. and 1x, = ~ 6 ksi. se Mohr's circle to determine the absolute maximum shear stress. lSTRATEGY: Use Mohr’s circle to determine the principal stresses ‘ and then determine the absolute maximum shear stress from the expressions (5-24).] U (ksi) ' §QLUTIQN: : . o- + 0' - ' The center of the circle is located at: ' 0c = ”LE—y = m = 7151' Q (5-5) 1,, = 0 The radius of the circle is: r = if (8131‘ ~ 'I'Irsz')2 + (— 6ksi)2 = 6.08ksr‘ 7- (ksi) _ The principal stresses are: 01 = cc + r = 7 ksi + 6.08 ksi = 13.08 ksi 02 = cc - r = 7 ksi - 6.08 ksi = 0.92 ksi ' Using equations (5.24) to determine the absolute maximum shear stress: 0.92ksi 0': ‘ 0'2 . ﬂ = 13.0813: = 6.5 4"”. g = =0_46,m- = W =3.04ksi , ANS 333m = 6.54 ksi T PRQBLEM §-§.g1 ‘I‘ For the state of plane stress ox= 240 MPa. 6,, = -120 MPa. and 1,, = 240 MPa, use Mohr's circle to determine the ‘1 absolute maximum shear stress. ‘. : The center of the circle is located at: 0c = 0'. + “r = W = 60MPa 2 2 1,, = 240 MPa The radius of the circle is: r = 1[(240MPa — 60MP0)2 + (240MHz)! = 300MPa " The principal stresses are: ' c1=ac+r =60 MPa+300 MPa=360 MPa 02 = cc - r = 60 MPa - 300 MPa = -240 MPa 1, The absolute maximum shear stress is: . ANS M = egg MPa 275 ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

HW 9 - PBQBLEM 5—3 s The components of plane stress at a...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online