hw6 - Problem 7.29 Determine the coordinates of the cen-...

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Unformatted text preview: Problem 7.29 Determine the coordinates of the cen- troicts. Solution: Divide the shape up into a rectangle, a semicircle, and a circular cutout as shown. Note that the y coordinates of the centroids of all three component areas lie on the :1; axis. Thus, y t O for the combined area. Rectangle: Areal = a.(2R) = 9600 mmg, andxl : a/2 : 60 mm. Scmt'circle: See example 7.3 or 7.4 for the value of the a: coordinate of the centroid of a semicircle. Also note the a: displacement of the centroid retative to the y axis. Areag : «HZ/2 : 2513 mmg, X2 = a. + (4R)/(31r) = 137.0 mm. Cutout: Areag : 7r'r'2 : 1257 mmz, X3 = 120 mm. Combined Area: x : (xlAreal + szreag M xaAreas)/(Area1 + Areag 7 Areag) : 70.9 mm Problem 7.30 Determine the coordinates of the cen— troids. Solution: The strategy is to find the centroid for the half circle area, and use the resuit in the composite algorithm. The area: The element of area is a vertical strip y high and aim wide. From the equation of the circle, y = iv R2 —- (1:3. The height of the strip will be twice the positive value. so that (1.4 = 2\/ R'2 — 2:2 dm, from which R A:f dA=2f (Hz—m2)1/2d:v A 0 :1: 1552—302 MgtT. The xvcaordinatc: R fwd/4:2] as 32—3:de A D W2 [_ (R2W32)3/2]R m 2R3 3 0 3 Divide by A: x = $3 The y-crmrdinate: From symmetry, the y—coordinate is zero. The composite: For a complete half cincle x1 : 4221?) r 8.488 LIL. For the inner half circie x2 : 4.244 in. The areas are A1 = 628.32 in.2 and A2 2 157.08 1112. TEA; "' 321‘ M Problem 7.49 Determine the reactions at A and B. Soiution: Let us break the load into two parts and me the ma anaiogy. For Loud L1 = EEdi—lgmfor (0 g m S L/Z) For Load Lg using the area analogy, toad L1 acts 2’3 of the distance from the origin to LIZ. Thus x1 : L / 3 Load 2 L L L2 3 f 0.10 (in: 2 wow : L/‘Z [1/2 And from the area analogy, L2 acts half way hetwaen U2 and L. _E x 2 4 mg};- Now we can find the support reactions é-‘m 3% Bow Solving the third eqn. Lwo 31M, 1, 11 B 3 fl 2 m y 6 Jr 4 12 From the second eqn, fly + By = ng0 W0 Hence Ay : glwu a By : wig“ AB = 0 A1, = —Lwo/6 By = 11 Luau/12 H Loan 2 N 3L 4 )=0 Problem 7.52 The distributed load is w = 6x + 0.4392 N/m. Determine the reactions at A and B. Solution: The total distributed and is 6 6 "'- 3 = .m3 F_f0 w( )dw ffl(em+o4 )da: 1:2 m4 6 0.4(36)2 F: 67 0.4—— = 3.3 w [ 2 + 4L [ 6+ 4 ] F = 237.6 N Using the area analogy, the point of application of the equivalent con— centrated force F is 6 5 f 3310(3) aim f (63:2 + 0.43:4) da: 0 7 o $ _ _—____m By = 263.5 N Problem 7.88 The bar has a mass of 80 kg. What are the reactions at A and B? Solution: Break the bar into two parts and and the masses and centers of masses of the two parts. The length of the bar is L ZL1+152=2m+21ITR/4(R=2m) L :2+7rrn m1 =31.12kg x1=1m mg 248.88kg x223.27m A320 Ay+Bymmlg—m2g=0 —x1mlg — xzmgg + 4131, = 0 Solving Am :0, Ag :316N, B:469N Problem 7.89 The semicircular part of the homoge- neous slender bar lies in the m m 2 plane. Determine the center of mass of the bar. Solution: The bar is divided into three segments plus the compos- ite. The lengths and the centroids are given in the table: The composite length is: 3 L :ZLi. i=1 The composite coordinates are: Problem 9.8 The prismatic bar has a solid circular Free Body Diagram: cross section with 30-min radius. It is suspended from - one end and is loaded only by its own weight. The mass (m 9 ) / A density of the homogeneous material is 2800 kgfma’. De— termine the average normal stress at the plane P, where m is the distance from the bottom of the bar in meters. Strategy: Draw a free—body diagram of the part of the bar below the plane P. Solution: The weight of the cylinder for any distance a: from its bottom is: W = pgvrr2x r (2800 kg/m3)(9.81 m/sec2}1r(0.03 m)2:c W = 77.66:]: N The stress produced in supporting this weight is: or = W/A : (77.66:: N)/1r(0.03 m)? ANS: a = 27,4673 Pa : 27.55.: kPa Problem 9.9 The beam has cross-sectional area A 2 0.0625 m2. What are average normal stress and the mag- nitude of the average shear stress at the plane P? Strategy: Draw the free-body diagram of the entire beam and determine the reactions at the pin and roller supports. Then determine the average normal and shear stresses by drawing the free-body diagram of the part of the beam to the left of plane P. Free Body Diagram: 2kN 5° \LB kN—m 4kN 03 A A TA O'A WA B X A l——> A)! B A A) F—Z m 2 m 2 m—-l Y yFZ m Solution: To find Am, sum horizontai forces. EFE=0=4kN—Am 145:4th To find AU, sum moments about point B: 2MB 2' 0 = (2 kN)(2 m) m (6 RN — m} + Am; 113) A9 = 333.33 N J, Efy =0=~333.3—2000+By =0 By : 2333.33 N T Now cut the beam through the plane P. Using the left-hand portion of the beam: 2pm = o = 0A M Am : aAVG(0.0625 1112) m 4,030 N ANS: ram/G = 64 kPa. Summing vertical forces on the leftrhand portion of the beam: 2s}, : n = —Ay + TAVGA : 433.3,? + mmmnsas m2) ANS: TAVG : 5.33 th-l Problem 9.16 The fixture shown connects a 50-mm diameter bridge cable to a flange that is attached to the bridge. A 60~mm diameter circular pin connects the fixture to the flange. If the average normal stress in the cable is crAV = 120 MP3,, what average shear stress my must the pin support? Solution: The average normal stress of 120 We in the cable means that the axial load in the cable is: P x 65Lng = (120 x 10“ N/m2)(1r)(0.025 m)2 P t 236 kN We see that the shear load in the pin is divided between two areas, one on either side of the flange. The average shear stress which must be supported by the pin is: Tug = P/(2A) = {236, 000 N)/(2)[1r(l).03 m2)] ANS: «rm = 41.7MPa. Problem 9.17 Consider the fixture shown in Prob— lem 9.16. The cable will safely support an average nor- mal stress of 700 MPa and the circular pin will safely support an average shear stress of 220 MPa. Based on these criteria, what is the largest tensile load the cable will safely support? Solution: Calculating the largest load which can be supported by the cable: Free Body Diagram: (FMAX)CABLE : (UAVG)ALLOWA : (700X106 N/m2){7r(0.025 1102i (FMAX)CABLE = 1.37 MN Calculating the largest load which can be supported by the pin: [FMAx)P1N 2 2KTAVG )MAxA] '= ZKZQOX 106 N/m2)(rr}(0.l}3 m)2] {mgrth = 1.24 MN We see that the cable can support a larger load than the pin. Any load larger than 1.24 MN will cause the pin to fail in shear. The largest load which can be supported by the cable and pin is: ANS: FMAX = 1.24 MN Problem 9.18 The truss is made of prismatic bars with cross-sectional area A = 0.25 £132. Determine the av- erage normal stress in member BE acting on a plane perpendicular to the axis of the member. Solution: We can determine the axial load in member BE directly by summing vertical forces on the FBD. EF-y = U = —8, 000 lb -— 10, 000 lb + PBE (sin 45°} PBE = 25.5 kip The average normal stress in member BE of the truss is: (:an = (PBEVA = (25, 500 lb) /(o.25 a2) ANS: 0an = 102 ksf = 703 psi Free Body Diagram: C Problem 9.34 A traction distribution :5 acts on a plane surface A. The value of t at a given point on A is t = 45i + 40j e 30k kPa. The unit vectori is perpendicular to A and points away from the material. What is the normal stress or at the given point? Solution: The i-componcnt of the applied stress is the only portion which con— tributes to NORMAL stress. Thej- and k—components act in the plane of the material and so contribute to shear stress. The normal stress at any point on the plane surface is: ANS: cravg = 45 kPn Problem‘9.35 In Problem 9.34, what is the magnitude of the shear stress r at the given point? Solution: The components of the applied stress which produce shear are the j- and k-componcnts. The resultant of these two orthogonal components is: R =1/t§+t:: (40 kPa)2 +(—301~;Pa)2 ANS: a : so kPa Problem 9.36 A traction distribution 1: acts on a plane surface A. The value of t at a given point on A is t : 3000i — 2000j + 6000k psi. The unit vector e = (5/7)i + (3/7)j + (2/7)}: is perpen- dicular to A and points away from the material. What is the normal stress a at the given point? Solution: To find that portion of the applied stress which is normal to the plane surface it is necessary to find the scalar product between the applied stress and the unit vector which is normal to the plane. Finding the scalar product between the vectors: oavg = t-e = (3000iw2000j+6000k}-{6/7i+3/7j+2/7k) = [(3000 psi)(6/7)]+[(72UUU psi)(3/7}]+[(6000 psi)(2/7)] ANS: am : 3430 psi Problem 9.37 A line of length dL at a particular point of a material in a reference state has length dL’ = 1.2 dL in a deformed state. What is the extensional strain cor— responding to that particular point and the direction of the line 01L? Solution: Extensional strain is the ratio of the change in length to the length in the reference state. a = (dL' A dL)/a!L : {1.2dL — dL)/dL a: 0.2 ANS: Problem 9.48 When the truss is subjected to the ver- tical force F, joint A moves a distance 1) : 0.3 m ver- tically and a distance u m 0.1 m horizontally. If the extensional strain 5A3 in the direction parallel to mem- ber AB is uniform throughout the length of the member, what is 6A3? Diagram: Solution: The reference length of member AB is: L: 2”” =2.309m sin 60° The deformed length of member AB is: L’ = (2.2996 mm)2 + (1.2545 111)2 = 2.6195 The strain for member AB is: L’—L _ 2‘9195 m—2.309 m 5A3 = L — 2.3o9 111 5,13 : 0.1344 Problem 9.49 In Problem 9.48, if the extensional strain 5A0 in the directional parallel to member AC is uniform throughout the length of the member, what is 5A0? Diagram: / ’ ' '\ l \//.__.._._.__.__.____i. \5/ l-G.‘¥ m Solution: The reference length of member AC is: L : t/(2 m)2 + (2 m)2 : 2.828m The deformed length of member AC is: L' = if (2.299 mm)2 + (1.899 m)2 2 2.932 m The strain for member AC is: E m L’wL _ 2.982m72.&28m ’ L m 2.828 m s = 0.0548 ...
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This note was uploaded on 10/12/2009 for the course MAE 210306210 taught by Professor chatterjee during the Fall '08 term at UCLA.

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hw6 - Problem 7.29 Determine the coordinates of the cen-...

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