FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES
|
|
|
- Rudolf Crawford
- 6 years ago
- Views:
Transcription
1 LECTURE Third Editio FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES A. J. Clark School of Egieerig Departmet of Civil ad Evirometal Egieerig Chapter 7.4 b Dr. Ibrahim A. Assakkaf SPRING 3 ENES Mechaics of Materials Departmet of Civil ad Evirometal Egieerig Uiversit of Marlad, College Park LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 1 The Stress Trasformatio Equatios for Plae Stress Eample ENES Assakkaf The stresses show i Figure 1a act at a poit o the free surface of a stressed bod. Determie the ormal stresses ad t ad the shearig stress t at this poit if the act o the rotated stress elemet show i Figure 1b.
2 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. The Stress Trasformatio Equatios for Plae Stress ENES Assakkaf Eample (cot d) 7MPa 4MPa t t Figure 1 t 1MPa 15 (a) (b) LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 3 The Stress Trasformatio Equatios for Plae Stress ENES Assakkaf Eample (cot d) The give values are as follows: = 1 MPa, = 7 MPa, θ = 15, θ = t = 15 t = + 4 MPa t t t
3 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 4 The Stress Trasformatio Equatios for Plae Stress Eample (cot d) Applig Eq. 1 for the give values = cos θ + si θ + = 1cos siθ cosθ ( 15) 7si ( 15) + (4) si( 15) cos( 15) ENES Assakkaf = MPa = MPa (Tesio) = cos θ + si θ + t = 1cos siθ cosθ ( 15) 7si ( 15) + (4) si( 15) cos( 15) = MPa = 86 MPa (comprssio) t LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 5 The Stress Trasformatio Equatios for Plae Stress Eample (cot d) t = ( ) siθ cosθ + ( cos θ si θ ) = ( 1 ( 7) ) si(15) cos(15) + 4( cos ( 15) si ( 15) ) = MPa t 7MPa 4MPa t t =86MPa t =19.64MPa ENES Assakkaf =5.98MPa 1MPa 15
4 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 6 Pricipal Stresses ad Maimum Pricipal Stresses ENES Assakkaf The trasformatio equatios (Eq. 1 or 13) provides a meas for determiig the ormal stress ad the shearig stress t o differet plaes through a poit O i stressed bod. Cosider, for eample, the state of stress at a poit O of the free surface of a structure or machie compoet (Fig. 13). LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 7 Pricipal Stresses ad Maimum Pricipal Stresses ENES Assakkaf Surfaces perpedicular to z-ais are stress-free. z 7 ksi 1ksi o 5ksi (a) Figure 13 (b)
5 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 8 Pricipal Stresses ad Maimum Pricipal Stresses ENES Assakkaf As the elemet is rotated through a agle θ about a ais perpedicular to the stressfree surface, the ormal stress ad the shearig stress t o differet plaes var cotiuousl as show i Figure 14. For desig purposes, critical stress at the poit are usuall the maimum tesile (or compressive) ad shearig stresses. LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 9 Pricipal Stresses ad Maimum Pricipal Stresses ENES Assakkaf The pricipal stresses are the maimum ormal stress ma ad miimum ormal stress mi. I geeral, these maimum ad miimum or pricipal stresses ca be determied b plottig curves similar to those of Fig. 14. But this process is time-cosumig, ad therefore, geeral methods are eeded.
6 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 1 Pricipal Stresses ad Maimum 4 3 ENES Assakkaf Variatio of Stresses as Fuctios of θ Stress s Stress tt t 7 ksi Stress (Ksi) t θ 1 ksi 5 ksi -1 - Agle θ (Degrees) Figure 14 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 11 Pricipal Stresses ad Maimum Pricipal Stresses Pricipal Stresses for Special Loadig Coditios: Bar uder aial load P = ad A Shaft uder Pure Bedig P A ENES Assakkaf ma ma = (14) T ma c ma = ma = (15) J
7 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 1 Pricipal Stresses ad Maimum Pricipal Stresses for Aiall Loaded Bar P ENES Assakkaf Figure 15 Origial Area, A θ Iclied Area, A P F LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 13 Pricipal Stresses ad Maimum Pricipal Stresses for Aiall Loaded A P Bar N = P cosθ V = Psiθ = A cosθ Figure 16a θ θ θ V N ENES Assakkaf N P cosθ P P = = = cos θ = A A A A cosθ ( 1+ cos θ )
8 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 14 Pricipal Stresses ad Maimum A Pricipal Stresses for Aiall Loaded Bar Figure 16b N P N = P cosθ V = Psiθ = A cosθ θ θ θ V ENES Assakkaf V P siθ P P = = = siθ cosθ = si θ A A A A cosθ LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 15 Pricipal Stresses ad Maimum Pricipal Stresses for Aiall Loaded Bar is maimum whe θ = or 18 is maimum whe θ = 45 or 135 Also ma ma = (16) Therefore P = ad A ma ma = P A ENES Assakkaf (17)
9 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 16 Pricipal Stresses ad Maimum Fig.17 t da cos α ENES Assakkaf Pricipal Stresses for Shaft uder Pure Torsio α t da (c) α da si α da α A (b) (a) LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 17 Pricipal Stresses ad Maimum ENES Assakkaf Pricipal Stresses for Shaft uder Pure Torsio = siα cosα si α = ( cos α si α ) cos α = t = (18) (19) t da cos α α t da α da T ma c ma = ma = () J da si α
10 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 18 Pricipal Stresses ad Maimum Developmet of Pricipal Stresses Equatios Recall Eq 13 t = si θ + cos θ ENES Assakkaf + = + cos θ + si θ (13a) (13b) LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 19 Pricipal Stresses ad Maimum Developmet of Pricipal Stresses Equatios or ENES Assakkaf Differetiatig the first equatio with respect to θ ad equate the result to zero, gives d + d = + cos θ + si θ dθ dθ set = ( ) si θ + cos θ = 1 ta θ = or = ta p θ p (1)
11 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. Pricipal Stresses ad Maimum Developmet of Pricipal Stresses Equatios ENES Assakkaf Substitutig the epressio for θ p ito Eq. 13a, ields + p1, p ± + = () Eq. 1 gives the two pricipal stresses i the -plae, ad the third stress p3 = z =. LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 1 Pricipal Stresses ad Maimum Pricipal Stresses Pricipal stresses ma ad mi ca be computed from + p1, p ± + = (a) where subscript p refers to the plaes of maimum ad miimum values of. ENES Assakkaf
12 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. Pricipal Stresses ad Maimum ENES Assakkaf Locatio of the Plae of Pricipal Stresses θ p = 1 1 ta (b) LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 3 Pricipal Stresses ad Maimum Notes o Pricipal Stresses Equatio 1. Eq. gives the agle θ p ad θ p + 9 betwee -plae (or -plae) ad the mutuall perpedicular plaes o which the pricipal stresses act.. Whe ta θ p is positive, θ p is positive, ad the rotatio is couterclockwise from the - ad-plaes to the plaes o which the two pricipal stresses act. ENES Assakkaf
13 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 4 Pricipal Stresses ad Maimum Notes o Pricipal Stresses Equatio ENES Assakkaf 3. Whe ta θ p is egative, θ p is egative, ad the rotatio is clockwise. 4. The shearig stress is zero o plaes eperiecig maimum ad miimum values of ormal stresses. 5. If oe or both of the pricipal stresses from Eq. is egative, the algebraic maimum stress ca have a smaller absolute value tha the miimum stress. LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 5 Pricipal Stresses ad Maimum ENES Assakkaf Developmet of Maimum Shearig Stress Equatio = si θ cos θ or Recall Eq 13b: t + d t d = si θ + cos θ dθ dθ = ta θ = set ( ) cos θ si θ = or θ = ta 1 (3)
14 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 6 Pricipal Stresses ad Maimum Developmet of Pricipal Shearig Stress Equatio ENES Assakkaf Substitutig the epressio for θ ito Eq. 13b, ields p ± + Eq. 4 gives the maimum i-plae shearig stress. = (4) LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 7 Pricipal Stresses ad Maimum ENES Assakkaf Maimum I-Plae Maimum i-plae shearig stress ca be computed from p = ± + (4a) where the subscript p refers to the plae of maimum i-plae shearig stress p.
15 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 8 Pricipal Stresses ad Maimum Locatio of the Plae of Maimum ENES Assakkaf θ = 1 1 ta (4b) LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 9 Pricipal Stresses ad Maimum Notes o Pricipal Stresses ad Maimum I-Plae Equatio 1. The two agles θ p ad θ differ b 9, therefore, θ p ad θ are 45 apart.. This meas that the plaes i which the maimum i-plae shearig stress occur are 45 from the pricipal plaes. ENES Assakkaf
16 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 3 Pricipal Stresses ad Maimum ENES Assakkaf Notes o Pricipal Stresses ad Maimum I-Plae Equatio 3. The directio of the maimum shearig stress ca be determied b drawig a wedge-shaped block with two sides parallel to the plaes havig the maimum ad miimum pricipal stresses, ad with the third side at a agle of 45. The directio of the maimum shearig stress must oppose the larger of the two pricipal stresses. LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 31 Pricipal Stresses ad Maimum θ p Fig.18 ENES Assakkaf Wedge-shaped Block > p1 P θ p ma p p1 ma p p
17 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 3 Pricipal Stresses ad Maimum Useful Relatioships ENES Assakkaf The maimum value of t is equal to oe half the differece betwee the two i-plae pricipal stresses, that is p1 p p = (5) For plae stress, the sum of the ormal stresses o a two orthogoal plaes through a poit i a bod is a costat or i ivariat. = + p1 + p (6) LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 33 Pricipal Stresses ad Maimum Useful Relatioships ENES Assakkaf Whe a state of plae eists, oe of the pricipal stresses is zero. If the values of p1 ad p from Eq. 5 have the same sig, the the third pricipal stress p3 equals zero, will be either the maimum or miimum ormal stresses. Three possibilities: ( )/, ma = ( ) /, ma = ( )/ ma = p1 p p1 p
18 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 34 Pricipal Stresses ad Maimum Eample 3 ENES Assakkaf Normal ad shearig stresses o horizotal ad vertical plaes through a poit i a structural member subjected to plae stress are show i Figure 19. Determie ad show o a sketch the pricipal ad maimum shearig stresses. LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 35 Pricipal Stresses ad Maimum Eample 3 (cot d) ENES Assakkaf Fig.19 4ksi 6ksi 1ksi The give values for use i Eqs. ad 4 are: = +1 ksi = - 4 ksi = - 6 ksi
19 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 36 Pricipal Stresses ad Maimum Eample 3 (cot d) Usig Eq. a for the give values: + p = ± ( 4) 1 ( 4) = ± + ( 6) = 4 ± 1 Therefore, = = + 14 ksi = 14 ksi (T) P1 p p3 = = = 4 1 = 6 ksi = 6 ksi (C) z ENES Assakkaf LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 37 Pricipal Stresses ad Maimum Eample 3 (cot d) Sice the p1 ad p have opposite sig, the maimum shearig stress is p 1 p 14 ( 6) ma = = = = + 1 ksi The locatio θ p of the pricipal stresses is computed from Eq. b θ p = 1 ta 1 1 ( 6) ( 4) = ENES Assakkaf
20 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 38 Pricipal Stresses ad Maimum Eample 3 (cot d) 4ksi 6ksi 6ksi = 1 ksi = p ma Sketch of pricipal ad ma shearig stresses 1ksi ksi 1 ksi 45 6 ksi + = 1 4 = = 4 14 ksi ENES Assakkaf ksi (T) LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 39 Pricipal Stresses ad Maimum Eample 4 Normal ad shearig stresses o horizotal ad vertical plaes through a poit i a structural member subjected to plae stress are show i Figure. Determie ad show o a sketch the pricipal ad maimum shearig stresses ENES Assakkaf
21 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 4 Pricipal Stresses ad Maimum Eample 4 (cot d) ENES Assakkaf Fig. 36MPa 7 MPa 4 MPa The give values for use i Eqs. ad 4 are: = +7 MPa = +36 MPa = - 4 MPa LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 41 Pricipal Stresses ad Maimum Eample 4 (cot d) Usig Eq. a for the give values: + p = ± ( + 36) 7 ( + 36) = ± + ( 4) = 54 ± 3 Therefore, = = + 84 ksi = 84 MPa (T) P1 p p3 = = = 54 3 = + 4 ksi = 4 MPa (T) z ENES Assakkaf
22 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 4 Pricipal Stresses ad Maimum Eample 4 (cot d) ENES Assakkaf Sice the p1 ad p have the same sig, the maimum shearig stress is ma = p = = = + 4 MPa The locatio θ p of the pricipal stresses is computed from Eq. b θ p = 1 ta 1 = 1 ta 1 ( 4) = LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 43 p Pricipal Stresses ad Maimum Eample 4 (cot d) 36 MPa 6ksi 4 MPa Sketch of pricipal ad ma shearig stresses 7 MPa 6.57 p1 p 84 4 p = = = 3 MPa = 4 MPa ma ma 3 MPa 54 MPa 45 4 MPa 4 MPa 45 p3 = 84 MPa 4 MPa = = = 54 ENES Assakkaf 84 MPa MPa (T)
23 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 44 Mohr s Circle for Plae Stress ENES Assakkaf Itroductio Mohr s circle is a pictorial or graphical iterpretatio of the trasformatio equatios for plae stress. The process ivolves the costructio of a circle i such a maer that the coordiates of each poit o the circle represet the ormal ad shearig stresses o oe plae through the stressed LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 45 Maimum ENES Assakkaf Maimum shearig stress occurs for = ave ma = R = + ta θ s = o Note : defies two agles separated b 9 ad o offset fromθ p b 45 + = ave =
24 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 46 Mohr s Circle for Plae Stress ENES Assakkaf Itroductio Poit, ad the agular positio of the radius to the poit gives the orietatio of the plae. The proof that ormal ad shearig compoets of stress o arbitrar plae through a poit ca be represeted as poits o a circle follows from Eqs. 13a ad 13b. LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 47 Mohr s Circle for Plae Stress ENES Assakkaf Plae Stresses usig Mohr s Circle Recall Eqs. 13a ad 13b, + = + cos θ + si θ (13a) t = si θ + cos θ (13a) Squarig both equatios, addig, ad simplifig gives + = t + + (7)
25 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 48 Mohr s Circle for Plae Stress ENES Assakkaf Plae Stresses usig Mohr s Circle The previous equatio is ideed a equatio of a circle i terms of the variable ad t. The circle is cetered o the the ais at a distace ( - )/ from the ais, ad the radius of the circle is give b R = + (8) LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 49 Mohr s Circle for Plae Stress ENES Assakkaf Plae Stresses usig Mohr s Circle Normal stresses are plotted as horizotal coordiates, with tesile stresses (positive) plotted to the right of the origi ad compressive stresses (egative) plotted to the left. Shearig stresses are plotted as vertical coordiates, with those tedig to produce a clockwise rotatio of the stress elemet
26 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 5 Mohr s Circle for Plae Stress ENES Assakkaf Plae Stresses usig Mohr s Circle plotted above the -ais, ad those tedig to produce couterclockwise rotatio of the stress elemet plotted below the -ais. Sig covetios for iterpretig the ormal ad shearig stresses will be provided, ad illustrated through eamples. LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 51 Mohr s Circle for Plae Stress ENES Assakkaf Plae Stresses usig Mohr s Circle Mohr s circle for a poit subjected to plae stress ca be draw whe stresses o two mutuall perpedicular plaes through the poit are kow. A θ A t θ
27 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 5 Mohr s Circle for Plae Stress ENES Assakkaf Mohr s Circle Fig.1 p H, ) ( p + C θ θ p t V, ) ( p1 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 53 Mohr s Circle for Plae Stress Mohr s Circle ENES Assakkaf Maimum shearig stress occurs for = ave ma = R = ta θ s = Note : defies two agles separated b 9 = offset fromθ b 45 ave p + = + o o ad
28 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 54 Mohr s Circle for Plae Stress ENES Assakkaf Plae Stresses usig Mohr s Circle Drawig Procedure for Mohr s Circle 1. Choose a set of - coordiate aes.. Idetif the stresses, ad = ad list them with proper sig. 3. Draw a set of -coordiate aes with ad positive to the right ad upward, respectivel. 4. Plot the poit (, - ) ad label it poit V (vertical plae). LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 55 Mohr s Circle for Plae Stress ENES Assakkaf Plae Stresses usig Mohr s Circle Drawig Procedure for Mohr s Circle (cot d) 5. Plot the poit (, ) ad label it poit H (horizotal plae). 6. Draw a lie betwee V ad H. This establishes the ceter ad the radius R of Mohr s circle. 7. Draw the circle. 8. A etesio of the radius betwee C ad V ca be idetified as the -ais or referece lie for the agle measuremets (I.e., θ =).
29 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 56 Mohr s Circle for Plae Stress ENES Assakkaf Sig Covetios I a give face of the stressed elemet, the shearig stresses that teds to rotate the elemet clockwise will be plotted above the -ais i the circle. I a give face of the stressed elemet, the shearig stresses that teds to rotate the elemet couterclockwise will be plotted below the -ais i the circle. LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 57 Mohr s Circle for Plae Stress ENES Assakkaf Sig Covetios Fig. The followig jigle ma be helpful i rememberig this covetios: I the kitche, the clock is above, ad the couter is below. Beer ad Johsto (199) (a) Clockwise Above (b) Couterclockwise Below
30 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 58 Mohr s Circle for Plae Stress ENES Assakkaf Poits of Iterests o Mohr s Circle 1. Poit D that provides the pricipal stress p1.. Poit E that gives the pricipal stress p. 3. Poit A that provides the maimum iplae shearig stress - p ad the accompaied ormal stress avg that acts o the plae. LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 59 Mohr s Circle for Plae Stress ENES Assakkaf With the phsical sigificace of Mohr s circle for plae stress established, it ma be applied with simple geometric cosideratios. Critical values are estimated graphicall or calculated. For a kow state of plae stress,, plot the poits X ad Y ad costruct the circle cetered at C. + ave = R = + The pricipal stresses are obtaied at A ad B. = ± R ma,mi ave ta θ p = The directio of rotatio of O to Oa is the same as CX to CA.
31 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 6 Mohr s Circle for Plae Stress ENES Assakkaf With Mohr s circle uiquel defied, the state of stress at other aes orietatios ma be depicted. For the state of stress at a agle θ with respect to the aes, costruct a ew diameter X Y at a agle θ with respect to XY. Normal ad shear stresses are obtaied from the coordiates X Y. LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 61 Mohr s Circle for Plae Stress Mohr s circle for cetric aial loadig: ENES Assakkaf P =, = = A P = = = A Mohr s circle for torsioal loadig: Tc Tc = = = = = = J J
32 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 6 Mohr s Circle for Plae Stress Eample 5 The stresses show i Figure 3 act at a poit o the free surface of a stressed bod. Use Mohr s circle to determie the ormal ad shearig stresses at this poit o the iclied plae AB show i the figure. 14 ksi ENES Assakkaf Fig.3 A 5 1 B ksi LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 63 Mohr s Circle for Plae Stress ENES Assakkaf Eample 5 (cot d) The give values for use i drawig Mohr s circle are: C R = 3 = p1 = ksi = p = 14 ksi z = p3 = θ = ta = R = C = = 17 ksi 14 R = radius = = 3 t = Rsi 45.4 = 3si 45.4 = = C R cos ( 45.4) = 17 3(.7) = ksi t ( ) ( ).13 ksi
33 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 64 Mohr s Circle for Plae Stress ENES Assakkaf Eample 6 For the state of plae stress show, (a) costruct Mohr s circle, determie (b) the pricipal plaes, (c) the pricipal stresses, (d) the maimum shearig stress ad the correspodig ormal stress. LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 65 Mohr s Circle for Plae Stress ENES Assakkaf Eample 6 (cot d) SOLUTION: Costructio of Mohr s circle + ( 5) + ( 1) ave = = = MPa CF = 5 = 3MPa FX = 4MPa R = CX = ( 3) + ( 4) = 5MPa
34 LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 66 Eample 6 (cot d) Pricipal plaes ad stresses ma = OA = OC + CA = + 5 ma = 7MPa ma = OB = OC BC = 5 ma = 3MPa ENES Assakkaf FX = CP θ p = 53.1 ta θ p = θ p = LECTURE. FAILURE CRITERIA: MOHR S CIRCLE AND PRINCIPAL STRESSES (7.4) Slide No. 67 Eample 6 (cot d) ENES Assakkaf Maimum shear stress θ s = θ p + 45 θ s = ma = R ma = 5 MPa = ave = MPa
Failure Theories Des Mach Elem Mech. Eng. Department Chulalongkorn University
Failure Theories Review stress trasformatio Failure theories for ductile materials Maimum-Shear-Stress Theor Distortio-Eerg Theor Coulomb-Mohr Theor Failure theories for brittle materials Maimum-Normal-Stress
SHAFTS: STATICALLY INDETERMINATE SHAFTS
LECTURE SHAFTS: STATICALLY INDETERMINATE SHAFTS Third Editio A.. Clark School of Egieerig Departmet of Civil ad Eviromet Egieerig 7 Chapter 3.6 by Dr. Ibrahim A. Assakkaf SPRING 2003 ENES 220 Mechaics
Lecture 7: Polar representation of complex numbers
Lecture 7: Polar represetatio of comple umbers See FLAP Module M3.1 Sectio.7 ad M3. Sectios 1 ad. 7.1 The Argad diagram I two dimesioal Cartesia coordiates (,), we are used to plottig the fuctio ( ) with
MECHANICS OF MATERIALS
CHAPTER MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Teas Tech Universit Transformations of Stress and Strain 006 The McGraw-Hill Companies,
Balancing. Rotating Components Examples of rotating components in a mechanism or a machine. (a)
alacig NOT COMPLETE Rotatig Compoets Examples of rotatig compoets i a mechaism or a machie. Figure 1: Examples of rotatig compoets: camshaft; crakshaft Sigle-Plae (Static) alace Cosider a rotatig shaft
NATIONAL SENIOR CERTIFICATE GRADE 12
NATIONAL SENIOR CERTIFICATE GRADE MATHEMATICS P FEBRUARY/MARCH 009 MARKS: 50 TIME: 3 hours This questio paper cosists of 0 pages, a iformatio sheet ad 3 diagram sheets. Please tur over Mathematics/P DoE/Feb.
SHAFTS: STATICALLY INDETERMINATE SHAFTS
SHAFTS: STATICALLY INDETERMINATE SHAFTS Up to this poit, the resses i a shaft has bee limited to shearig resses. This due to the fact that the selectio of the elemet uder udy was orieted i such a way that
R is a scalar defined as follows:
Math 8. Notes o Dot Product, Cross Product, Plaes, Area, ad Volumes This lecture focuses primarily o the dot product ad its may applicatios, especially i the measuremet of agles ad scalar projectio ad
We will conclude the chapter with the study a few methods and techniques which are useful
Chapter : Coordiate geometry: I this chapter we will lear about the mai priciples of graphig i a dimesioal (D) Cartesia system of coordiates. We will focus o drawig lies ad the characteristics of the graphs
MECHANICS OF MATERIALS
00 The McGraw-Hill Companies, Inc. All rights reserved. T Edition CHAPTER MECHANICS OF MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Teas Tech Universit
CALCULUS BASIC SUMMER REVIEW
CALCULUS BASIC SUMMER REVIEW NAME rise y y y Slope of a o vertical lie: m ru Poit Slope Equatio: y y m( ) The slope is m ad a poit o your lie is, ). ( y Slope-Itercept Equatio: y m b slope= m y-itercept=
Fundamental Concepts: Surfaces and Curves
UNDAMENTAL CONCEPTS: SURACES AND CURVES CHAPTER udametal Cocepts: Surfaces ad Curves. INTRODUCTION This chapter describes two geometrical objects, vi., surfaces ad curves because the pla a ver importat
SOLID MECHANICS TUTORIAL BALANCING OF RECIPROCATING MACHINERY
SOLID MECHANICS TUTORIAL BALANCING OF RECIPROCATING MACHINERY This work covers elemets of the syllabus for the Egieerig Coucil Exam D5 Dyamics of Mechaical Systems. O completio of this tutorial you should
Calculus I Practice Test Problems for Chapter 5 Page 1 of 9
Calculus I Practice Test Problems for Chapter 5 Page of 9 This is a set of practice test problems for Chapter 5. This is i o way a iclusive set of problems there ca be other types of problems o the actual
U8L1: Sec Equations of Lines in R 2
MCVU U8L: Sec. 8.9. Equatios of Lies i R Review of Equatios of a Straight Lie (-D) Cosider the lie passig through A (-,) with slope, as show i the diagram below. I poit slope form, the equatio of the lie
Mathematics Extension 2
009 HIGHER SCHOOL CERTIFICATE EXAMINATION Mathematics Etesio Geeral Istructios Readig time 5 miutes Workig time hours Write usig black or blue pe Board-approved calculators may be used A table of stadard
Appendix F: Complex Numbers
Appedix F Complex Numbers F1 Appedix F: Complex Numbers Use the imagiary uit i to write complex umbers, ad to add, subtract, ad multiply complex umbers. Fid complex solutios of quadratic equatios. Write
MTH Assignment 1 : Real Numbers, Sequences
MTH -26 Assigmet : Real Numbers, Sequeces. Fid the supremum of the set { m m+ : N, m Z}. 2. Let A be a o-empty subset of R ad α R. Show that α = supa if ad oly if α is ot a upper boud of A but α + is a
Notes The Incremental Motion Model:
The Icremetal Motio Model: The Jacobia Matrix I the forward kiematics model, we saw that it was possible to relate joit agles θ, to the cofiguratio of the robot ed effector T I this sectio, we will see
Coordinate Systems. Things to think about:
Coordiate Sstems There are 3 coordiate sstems that a compter graphics programmer is most cocered with: the Object Coordiate Sstem (OCS), the World Coordiate Sstem (WCS), ad the Camera Coordiate Sstem (CCS).
NATIONAL SENIOR CERTIFICATE GRADE 12
NATIONAL SENIOR CERTIFICATE GRADE 1 MATHEMATICS P NOVEMBER 01 MARKS: 150 TIME: 3 hours This questio paper cosists of 13 pages, 1 diagram sheet ad 1 iformatio sheet. Please tur over Mathematics/P DBE/November
STRAIGHT LINES & PLANES
STRAIGHT LINES & PLANES PARAMETRIC EQUATIONS OF LINES The lie "L" is parallel to the directio vector "v". A fixed poit: "( a, b, c) " o the lie is give. Positio vectors are draw from the origi to the fixed
APPENDIX F Complex Numbers
APPENDIX F Complex Numbers Operatios with Complex Numbers Complex Solutios of Quadratic Equatios Polar Form of a Complex Number Powers ad Roots of Complex Numbers Operatios with Complex Numbers Some equatios
Chapter 2 Feedback Control Theory Continued
Chapter Feedback Cotrol Theor Cotiued. Itroductio I the previous chapter, the respose characteristic of simple first ad secod order trasfer fuctios were studied. It was show that first order trasfer fuctio,
Assignment 1 : Real Numbers, Sequences. for n 1. Show that (x n ) converges. Further, by observing that x n+2 + x n+1
Assigmet : Real Numbers, Sequeces. Let A be a o-empty subset of R ad α R. Show that α = supa if ad oly if α is ot a upper boud of A but α + is a upper boud of A for every N. 2. Let y (, ) ad x (, ). Evaluate
PHYSICS 116A Homework 2 Solutions
PHYSICS 6A Homework 2 Solutios I. [optioal] Boas, Ch., 6, Qu. 30 (proof of the ratio test). Just follow the hits. If ρ, the ratio of succcessive terms for is less tha, the hits show that the terms of the
A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece,, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet as
Sequences A sequence of numbers is a function whose domain is the positive integers. We can see that the sequence
Sequeces A sequece of umbers is a fuctio whose domai is the positive itegers. We ca see that the sequece 1, 1, 2, 2, 3, 3,... is a fuctio from the positive itegers whe we write the first sequece elemet
Solutions to quizzes Math Spring 2007
to quizzes Math 4- Sprig 7 Name: Sectio:. Quiz a) x + x dx b) l x dx a) x + dx x x / + x / dx (/3)x 3/ + x / + c. b) Set u l x, dv dx. The du /x ad v x. By Itegratio by Parts, x(/x)dx x l x x + c. l x
AP Calculus BC Review Applications of Derivatives (Chapter 4) and f,
AP alculus B Review Applicatios of Derivatives (hapter ) Thigs to Kow ad Be Able to Do Defiitios of the followig i terms of derivatives, ad how to fid them: critical poit, global miima/maima, local (relative)
ES.182A Topic 40 Notes Jeremy Orloff
ES.182A opic 4 Notes Jeremy Orloff 4 Flux: ormal form of Gree s theorem Gree s theorem i flux form is formally equivalet to our previous versio where the lie itegral was iterpreted as work. Here we will
Honors Calculus Homework 13 Solutions, due 12/8/5
Hoors Calculus Homework Solutios, due /8/5 Questio Let a regio R i the plae be bouded by the curves y = 5 ad = 5y y. Sketch the regio R. The two curves meet where both equatios hold at oce, so where: y
NATIONAL SENIOR CERTIFICATE GRADE 12
NATIONAL SENIOR CERTIFICATE GRADE MATHEMATICS P EXEMPLAR 008 MARKS: 50 TIME: 3 hours This questio paper cosists of pages, diagram sheets ad a formula sheet. Please tur over Mathematics/P DoE/Eemplar 008
U8L1: Sec Equations of Lines in R 2
MCVU Thursda Ma, Review of Equatios of a Straight Lie (-D) U8L Sec. 8.9. Equatios of Lies i R Cosider the lie passig through A (-,) with slope, as show i the diagram below. I poit slope form, the equatio
Chimica Inorganica 3
himica Iorgaica Irreducible Represetatios ad haracter Tables Rather tha usig geometrical operatios, it is ofte much more coveiet to employ a ew set of group elemets which are matrices ad to make the rule
a is some real number (called the coefficient) other
Precalculus Notes for Sectio.1 Liear/Quadratic Fuctios ad Modelig http://www.schooltube.com/video/77e0a939a3344194bb4f Defiitios: A moomial is a term of the form tha zero ad is a oegative iteger. a where
Problem Cosider the curve give parametrically as x = si t ad y = + cos t for» t» ß: (a) Describe the path this traverses: Where does it start (whe t =
Mathematics Summer Wilso Fial Exam August 8, ANSWERS Problem 1 (a) Fid the solutio to y +x y = e x x that satisfies y() = 5 : This is already i the form we used for a first order liear differetial equatio,
NATIONAL SENIOR CERTIFICATE GRADE 12
NATIONAL SENIOR CERTIFICATE GRADE MATHEMATICS P FEBRUARY/MARCH 03 MARKS: 50 TIME: 3 hours This questio paper cosists of pages, 3 diagram sheets ad iformatio sheet. Please tur over Mathematics/P DBE/Feb.
September 2012 C1 Note. C1 Notes (Edexcel) Copyright - For AS, A2 notes and IGCSE / GCSE worksheets 1
September 0 s (Edecel) Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright www.pgmaths.co.uk - For AS, A otes ad IGCSE / GCSE worksheets September 0 Copyright
EVALUATION OF GLASS FIBER/EPOXY INTERFACIAL STRENGTH BY THE CRUCIFORM SPECIMEN METHOD
EVALUATION OF GLASS FIBER/EPOX INTERFACIAL STRENGTH B THE CRUCIFORM SPECIMEN METHOD Ju KOANAGI, Hajime KATO, Akihiro KASHIMA, uichi IGARASHI, Keichi WATANABE 3, Ichiro UENO 4 ad Shiji OGIHARA 4 Istitute
REFLECTION AND REFRACTION
RFLCTON AND RFRACTON We ext ivestigate what happes whe a light ray movig i oe medium ecouters aother medium, i.e. the pheomea of reflectio ad refractio. We cosider a plae M wave strikig a plae iterface
A Recurrence Formula for Packing Hyper-Spheres
A Recurrece Formula for Packig Hyper-Spheres DokeyFt. Itroductio We cosider packig of -D hyper-spheres of uit diameter aroud a similar sphere. The kissig spheres ad the kerel sphere form cells of equilateral
Chapter 4. Fourier Series
Chapter 4. Fourier Series At this poit we are ready to ow cosider the caoical equatios. Cosider, for eample the heat equatio u t = u, < (4.) subject to u(, ) = si, u(, t) = u(, t) =. (4.) Here,
Complex Analysis Spring 2001 Homework I Solution
Complex Aalysis Sprig 2001 Homework I Solutio 1. Coway, Chapter 1, sectio 3, problem 3. Describe the set of poits satisfyig the equatio z a z + a = 2c, where c > 0 ad a R. To begi, we see from the triagle
Math 21C Brian Osserman Practice Exam 2
Math 1C Bria Osserma Practice Exam 1 (15 pts.) Determie the radius ad iterval of covergece of the power series (x ) +1. First we use the root test to determie for which values of x the series coverges
MOCK TEST - 02 COMMON ENTRANCE TEST 2012 SUBJECT: MATHEMATICS Time: 1.10Hrs Max. Marks 60 Questions 60. then x 2 =
MOCK TEST - 0 COMMON ENTRANCE TEST 0 SUBJECT: MATHEMATICS Time:.0Hrs Max. Marks 60 Questios 60. The value of si cot si 3 cos sec + + 4 4 a) 0 b) c) 4 6 + x x. If Ta - α + x + x the x a) cos α b) Taα c)
REVISION SHEET FP1 (MEI) ALGEBRA. Identities In mathematics, an identity is a statement which is true for all values of the variables it contains.
The mai ideas are: Idetities REVISION SHEET FP (MEI) ALGEBRA Before the exam you should kow: If a expressio is a idetity the it is true for all values of the variable it cotais The relatioships betwee
CHAPTER 8 SYSTEMS OF PARTICLES
CHAPTER 8 SYSTES OF PARTICLES CHAPTER 8 COLLISIONS 45 8. CENTER OF ASS The ceter of mass of a system of particles or a rigid body is the poit at which all of the mass are cosidered to be cocetrated there
Math 21B-B - Homework Set 2
Math B-B - Homework Set Sectio 5.:. a) lim P k= c k c k ) x k, where P is a partitio of [, 5. x x ) dx b) lim P k= 4 ck x k, where P is a partitio of [,. 4 x dx c) lim P k= ta c k ) x k, where P is a partitio
Topic 5 [434 marks] (i) Find the range of values of n for which. (ii) Write down the value of x dx in terms of n, when it does exist.
![Topic 5 [434 marks] (i) Find the range of values of n for which. (ii) Write down the value of x dx in terms of n, when it does exist.](/thumbs/90/102216025.jpg "Topic 5 [434 marks] (i) Find the range of values of n for which. (ii) Write down the value of x dx in terms of n, when it does exist.")
Topic 5 [44 marks] 1a (i) Fid the rage of values of for which eists 1 Write dow the value of i terms of 1, whe it does eist Fid the solutio to the differetial equatio 1b give that y = 1 whe = π (cos si
Mathematics Extension 2
004 HIGHER SCHOOL CERTIFICATE EXAMINATION Mathematics Etesio Geeral Istructios Readig time 5 miutes Workig time hours Write usig black or blue pe Board-approved calculators may be used A table of stadard
Chapter 10: Power Series
Chapter : Power Series 57 Chapter Overview: Power Series The reaso series are part of a Calculus course is that there are fuctios which caot be itegrated. All power series, though, ca be itegrated because
The z-transform. 7.1 Introduction. 7.2 The z-transform Derivation of the z-transform: x[n] = z n LTI system, h[n] z = re j
![The z-transform. 7.1 Introduction. 7.2 The z-transform Derivation of the z-transform: x[n] = z n LTI system, h[n] z = re j](/thumbs/82/85448534.jpg "The z-transform. 7.1 Introduction. 7.2 The z-transform Derivation of the z-transform: x[n] = z n LTI system, h[n] z = re j")
The -Trasform 7. Itroductio Geeralie the complex siusoidal represetatio offered by DTFT to a represetatio of complex expoetial sigals. Obtai more geeral characteristics for discrete-time LTI systems. 7.
Systems of Particles: Angular Momentum and Work Energy Principle
1 2.003J/1.053J Dyamics ad Cotrol I, Sprig 2007 Professor Thomas Peacock 2/20/2007 Lecture 4 Systems of Particles: Agular Mometum ad Work Eergy Priciple Systems of Particles Agular Mometum (cotiued) τ
Implicit function theorem
Jovo Jaric Implicit fuctio theorem The reader kows that the equatio of a curve i the x - plae ca be expressed F x, =., this does ot ecessaril represet a fuctio. Take, for example F x, = 2x x =. (1 either
Chapter 1. Complex Numbers. Dr. Pulak Sahoo
Chapter 1 Complex Numbers BY Dr. Pulak Sahoo Assistat Professor Departmet of Mathematics Uiversity Of Kalyai West Begal, Idia E-mail : sahoopulak1@gmail.com 1 Module-2: Stereographic Projectio 1 Euler
Quiz. Use either the RATIO or ROOT TEST to determine whether the series is convergent or not.
Quiz. Use either the RATIO or ROOT TEST to determie whether the series is coverget or ot. e .6 POWER SERIES Defiitio. A power series i about is a series of the form c 0 c a c a... c a... a 0 c a where
Review Problems Math 122 Midterm Exam Midterm covers App. G, B, H1, H2, Sec , 8.9,
Review Problems Math Midterm Exam Midterm covers App. G, B, H, H, Sec 8. - 8.7, 8.9, 9.-9.7 Review the Cocept Check problems: Page 6/ -, Page 690/- 0 PART I: True-False Problems Ch. 8. Page 6 True-False
AIEEE 2004 (MATHEMATICS)
AIEEE 00 (MATHEMATICS) Importat Istructios: i) The test is of hours duratio. ii) The test cosists of 75 questios. iii) The maimum marks are 5. iv) For each correct aswer you will get marks ad for a wrog
Apply change-of-basis formula to rewrite x as a linear combination of eigenvectors v j.
Eigevalue-Eigevector Istructor: Nam Su Wag eigemcd Ay vector i real Euclidea space of dimesio ca be uiquely epressed as a liear combiatio of liearly idepedet vectors (ie, basis) g j, j,,, α g α g α g α
MATH CALCULUS II Objectives and Notes for Test 4
MATH 44 - CALCULUS II Objectives ad Notes for Test 4 To do well o this test, ou should be able to work the followig tpes of problems. Fid a power series represetatio for a fuctio ad determie the radius
VTU EDUSAT PROGRAMME-17. DYNAMICS OF MACHINES Subject Code -10 ME 54 BALANCING OF RECIPROCATING MASSES ADARSHA H G
Subject Code -1 ME 5 BALANCING OF RECIPROCATING MASSES Notes Compiled by: ASSOCIATE PROFESSOR DEPARTMENT OF MECHANICAL ENGINEERING MALNAD COLLEGE BALANCING OF ENGINEERING HASSAN -57 OF. KARNATAKA RECIPROCATING
Exponential and Trigonometric Functions Lesson #1
Epoetial ad Trigoometric Fuctios Lesso # Itroductio To Epoetial Fuctios Cosider a populatio of 00 mice which is growig uder plague coditios. If the mouse populatio doubles each week we ca costruct a table
Summary: CORRELATION & LINEAR REGRESSION. GC. Students are advised to refer to lecture notes for the GC operations to obtain scatter diagram.
Key Cocepts: 1) Sketchig of scatter diagram The scatter diagram of bivariate (i.e. cotaiig two variables) data ca be easily obtaied usig GC. Studets are advised to refer to lecture otes for the GC operatios
Two or more points can be used to describe a rigid body. This will eliminate the need to define rotational coordinates for the body!
OINTCOORDINATE FORMULATION Two or more poits ca be used to describe a rigid body. This will elimiate the eed to defie rotatioal coordiates for the body i z r i i, j r j j rimary oits: The coordiates of
Consortium of Medical Engineering and Dental Colleges of Karnataka (COMEDK) Undergraduate Entrance Test(UGET) Maths-2012
Cosortium of Medical Egieerig ad Detal Colleges of Karataka (COMEDK) Udergraduate Etrace Test(UGET) Maths-0. If the area of the circle 7 7 7 k 0 is sq. uits, the the value of k is As: (b) b) 0 7 K 0 c)
For example suppose we divide the interval [0,2] into 5 equal subintervals of length
![For example suppose we divide the interval [0,2] into 5 equal subintervals of length](/thumbs/83/87992504.jpg "For example suppose we divide the interval [0,2] into 5 equal subintervals of length")
Math 1206 Calculus Sec 1: Estimatig with Fiite Sums Abbreviatios: wrt with respect to! for all! there exists! therefore Def defiitio Th m Theorem sol solutio! perpedicular iff or! if ad oly if pt poit
JEE ADVANCED 2013 PAPER 1 MATHEMATICS
Oly Oe Optio Correct Type JEE ADVANCED 0 PAPER MATHEMATICS This sectio cotais TEN questios. Each has FOUR optios (A), (B), (C) ad (D) out of which ONLY ONE is correct.. The value of (A) 5 (C) 4 cot cot
NATIONAL SENIOR CERTIFICATE GRADE 12
NATIONAL SENIOR CERTIFICATE GRADE 1 MATHEMATICS P FEBRUARY/MARCH 014 MARKS: 150 TIME: 3 hours This questio paper cosists of 1 pages, 3 diagram sheets ad 1 iformatio sheet. Please tur over Mathematics/P
MATHEMATICS 9740 (HIGHER 2)
VICTORIA JUNIOR COLLEGE PROMOTIONAL EXAMINATION MATHEMATICS 970 (HIGHER ) Frida 6 Sept 0 8am -am hours Additioal materials: Aswer Paper List of Formulae (MF5) READ THESE INSTRUCTIONS FIRST Write our ame
Continuous Functions
Cotiuous Fuctios Q What does it mea for a fuctio to be cotiuous at a poit? Aswer- I mathematics, we have a defiitio that cosists of three cocepts that are liked i a special way Cosider the followig defiitio
Principle Of Superposition
ecture 5: PREIMINRY CONCEP O RUCUR NYI Priciple Of uperpositio Mathematically, the priciple of superpositio is stated as ( a ) G( a ) G( ) G a a or for a liear structural system, the respose at a give
ECE Spring Prof. David R. Jackson ECE Dept. Notes 20
ECE 6341 Sprig 016 Prof. David R. Jackso ECE Dept. Notes 0 1 Spherical Wave Fuctios Cosider solvig ψ + k ψ = 0 i spherical coordiates z φ θ r y x Spherical Wave Fuctios (cot.) I spherical coordiates we
Section 1.1. Calculus: Areas And Tangents. Difference Equations to Differential Equations
Differece Equatios to Differetial Equatios Sectio. Calculus: Areas Ad Tagets The study of calculus begis with questios about chage. What happes to the velocity of a swigig pedulum as its positio chages?
Upper bound for ropelength of pretzel knots
Upper boud for ropelegth of pretzel kots Safiya Mora August 25, 2006 Abstract A model of the pretzel kot is described. A method for predictig the ropelegth of pretzel kots is give. A upper boud for the
6. If si () + cos () =, the is equal to :. If < REVISION SHEET FP1 (MEI) ALGEBRA. Identities In mathematics, an identity is a statement which is true for all values of the variables it contains.
the Further Mathematics etwork wwwfmetworkorguk V 07 The mai ideas are: Idetities REVISION SHEET FP (MEI) ALGEBRA Before the exam you should kow: If a expressio is a idetity the it is true for all values
1 1 2 = show that: over variables x and y. [2 marks] Write down necessary conditions involving first and second-order partial derivatives for ( x0, y
![1 1 2 = show that: over variables x and y. [2 marks] Write down necessary conditions involving first and second-order partial derivatives for ( x0, y](/thumbs/95/125864248.jpg "1 1 2 = show that: over variables x and y. [2 marks] Write down necessary conditions involving first and second-order partial derivatives for ( x0, y")
Questio (a) A square matrix A= A is called positive defiite if the quadratic form waw > 0 for every o-zero vector w [Note: Here (.) deotes the traspose of a matrix or a vector]. Let 0 A = 0 = show that:
Chapter 13: Complex Numbers
Sectios 13.1 & 13.2 Comple umbers ad comple plae Comple cojugate Modulus of a comple umber 1. Comple umbers Comple umbers are of the form z = + iy,, y R, i 2 = 1. I the above defiitio, is the real part
Chapter 2 The Solution of Numerical Algebraic and Transcendental Equations
Chapter The Solutio of Numerical Algebraic ad Trascedetal Equatios Itroductio I this chapter we shall discuss some umerical methods for solvig algebraic ad trascedetal equatios. The equatio f( is said
Section 7. Gaussian Reduction
7- Sectio 7 Gaussia eductio Paraxial aytrace Equatios eractio occurs at a iterace betwee two optical spaces. The traser distace t' allows the ray height y' to be determied at ay plae withi a optical space
6.003 Homework #3 Solutions
6.00 Homework # Solutios Problems. Complex umbers a. Evaluate the real ad imagiary parts of j j. π/ Real part = Imagiary part = 0 e Euler s formula says that j = e jπ/, so jπ/ j π/ j j = e = e. Thus the
Measuring Scales. Measuring Scales
Measurig Scales To measure a legth, a metre scale is geerally used, which is graduated to cetimeter ad millimeter, ad is oe metre i legth. For the measuremet of a legth with a metre scale we adopt the
Session 5. (1) Principal component analysis and Karhunen-Loève transformation
200 Autum semester Patter Iformatio Processig Topic 2 Image compressio by orthogoal trasformatio Sessio 5 () Pricipal compoet aalysis ad Karhue-Loève trasformatio Topic 2 of this course explais the image
4.1 Sigma Notation and Riemann Sums
0 the itegral. Sigma Notatio ad Riema Sums Oe strategy for calculatig the area of a regio is to cut the regio ito simple shapes, calculate the area of each simple shape, ad the add these smaller areas
Elementary Linear Algebra
Elemetary Liear Algebra Ato & Rorres th Editio Lectre Set Chapter : Eclidea Vector Spaces Chapter Cotet Vectors i -Space -Space ad -Space Norm Distace i R ad Dot Prodct Orthogoality Geometry of Liear Systems
MTH112 Trigonometry 2 2 2, 2. 5π 6. cscθ = 1 sinθ = r y. secθ = 1 cosθ = r x. cotθ = 1 tanθ = cosθ. central angle time. = θ t.
MTH Trigoometry,, 5, 50 5 0 y 90 0, 5 0,, 80 0 0 0 (, 0) x, 7, 0 5 5 0, 00 5 5 0 7,,, Defiitios: siθ = opp. hyp. = y r cosθ = adj. hyp. = x r taθ = opp. adj. = siθ cosθ = y x cscθ = siθ = r y secθ = cosθ
Most text will write ordinary derivatives using either Leibniz notation 2 3. y + 5y= e and y y. xx tt t
Itroductio to Differetial Equatios Defiitios ad Termiolog Differetial Equatio: A equatio cotaiig the derivatives of oe or more depedet variables, with respect to oe or more idepedet variables, is said
SEQUENCE AND SERIES NCERT
9. Overview By a sequece, we mea a arragemet of umbers i a defiite order accordig to some rule. We deote the terms of a sequece by a, a,..., etc., the subscript deotes the positio of the term. I view of
SAFE HANDS & IIT-ian's PACE EDT-10 (JEE) SOLUTIONS
. If their mea positios coicide with each other, maimum separatio will be A. Now from phasor diagram, we ca clearly see the phase differece. SAFE HANDS & IIT-ia's PACE ad Aswer : Optio (4) 5. Aswer : Optio
6 Integers Modulo n. integer k can be written as k = qn + r, with q,r, 0 r b. So any integer.
6 Itegers Modulo I Example 2.3(e), we have defied the cogruece of two itegers a,b with respect to a modulus. Let us recall that a b (mod ) meas a b. We have proved that cogruece is a equivalece relatio
(c) Write, but do not evaluate, an integral expression for the volume of the solid generated when R is
Calculus BC Fial Review Name: Revised 7 EXAM Date: Tuesday, May 9 Remiders:. Put ew batteries i your calculator. Make sure your calculator is i RADIAN mode.. Get a good ight s sleep. Eat breakfast. Brig:
SINGLE CORRECT ANSWER TYPE QUESTIONS: TRIGONOMETRY 2 2
Class-Jr.X_E-E SIMPLE HOLIDAY PACKAGE CLASS-IX MATHEMATICS SUB BATCH : E-E SINGLE CORRECT ANSWER TYPE QUESTIONS: TRIGONOMETRY. siθ+cosθ + siθ cosθ = ) ) ). If a cos q, y bsi q, the a y b ) ) ). The value
NAME: ALGEBRA 350 BLOCK 7. Simplifying Radicals Packet PART 1: ROOTS
NAME: ALGEBRA 50 BLOCK 7 DATE: Simplifyig Radicals Packet PART 1: ROOTS READ: A square root of a umber b is a solutio of the equatio x = b. Every positive umber b has two square roots, deoted b ad b or
CS 112 Transformations II. Slide 1
CS 112 Trasformatios II Slide 1 Compositio of Trasformatios Example: A poit P is first traslated ad the rotated. Traslatio matrix T, Rotatio Matrix R. After Traslatio: P = TP After Rotatio: P =RP =RTP
Curve Sketching Handout #5 Topic Interpretation Rational Functions
Curve Sketchig Hadout #5 Topic Iterpretatio Ratioal Fuctios A ratioal fuctio is a fuctio f that is a quotiet of two polyomials. I other words, p ( ) ( ) f is a ratioal fuctio if p ( ) ad q ( ) are polyomials
The Minimum Distance Energy for Polygonal Unknots
The Miimum Distace Eergy for Polygoal Ukots By:Johaa Tam Advisor: Rollad Trapp Abstract This paper ivestigates the eergy U MD of polygoal ukots It provides equatios for fidig the eergy for ay plaar regular
Bivariate Sample Statistics Geog 210C Introduction to Spatial Data Analysis. Chris Funk. Lecture 7
Bivariate Sample Statistics Geog 210C Itroductio to Spatial Data Aalysis Chris Fuk Lecture 7 Overview Real statistical applicatio: Remote moitorig of east Africa log rais Lead up to Lab 5-6 Review of bivariate/multivariate
Complex Numbers. Brief Notes. z = a + bi
Defiitios Complex Numbers Brief Notes A complex umber z is a expressio of the form: z = a + bi where a ad b are real umbers ad i is thought of as 1. We call a the real part of z, writte Re(z), ad b the
Problem 1. Problem Engineering Dynamics Problem Set 9--Solution. Find the equation of motion for the system shown with respect to:
2.003 Egieerig Dyamics Problem Set 9--Solutio Problem 1 Fid the equatio of motio for the system show with respect to: a) Zero sprig force positio. Draw the appropriate free body diagram. b) Static equilibrium