BME 2220 Homework 2 _2_

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BME 2220 Homework 2 Due February 17, 2022 Problem 1 During a match, a wrestler pushes on the head of his opponent with force F=1200 N, as shown in the picture on the left. His opponent holds his head in equilibrium by applying an isometric contraction force in his neck muscle (M). The musculature in the neck is very complex anatomically, but for the sake of this problem, assume the neck muscles can be simulated as a single cable. Treat the connection of the head to the C1 vertebrae as a hinge (point C). By chance, the lines of action of F, M, and Cy all happen to pass through point A in this problem. (Note that the lack of dimensions in this problem is intentional. They aren’t needed to solve the problem.) Using the free-body diagram provided on the right, determine the magnitude of the force in the neck muscles (M), as well as the reaction forces at the neck joint (Cx and Cy), needed for the opponent to hold his head in equilibrium. (Units: N) Problem 2 Suppose you have been asked to find the Young’s modulus of an unknown material. The geometry of the material sample is shown below. It has an original length L = 8 cm, and a perfectly square hollow cross-section, with dimensions shown on the right. When you apply a force F = 150 N to the sample as shown below, the sample deforms to a new length L new =8.10 cm. a. Calculate Young’s Modulus (units: N/cm2). b. Calculate the normal and shear stress on the angled plane drawn in the figure. Ɵ = 65°. Ignore weight (units: N/cm 2 ) F = 1200 N 65° F = 150 N F = 150 N cut Ɵ
EMA D Ncos Gs X FUSCO X Mcoscos Fios 30 m t Fx Fios 30 M cos 65 Cx D Cx 0 spy Cy Fs ing o t Msm 65 12008 to t 2459.029 s co Cy 2828 627 N Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com
Strain E dy du o l am 1 8 am E of em 0.0125cm stress o Fia pg qq.gg At 2.55 62 2 as am Young modulus EE Ole 1 506 5333.33 Mom Normal Stress O V Rosa Rason sheer stress E Ao oh since fo T Rosa no sing Its Ing Alsina It cososino a n II.is s I 2 4826 Area bh 2.4826 2 150 N as 54.75 2 151 coscossincos 2 vs Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com
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Problem 3 A frame with negligible mass is supported by a hinge at A and a rope (cable connections at “B” and “D”) that passes over a smooth pulley at C, as shown to the right. Determine the force in the cable and the reaction force(s) at A (units: N). Your F.B.D. will be of the frame. Problem 4 Given the beam below: a. Draw a free body diagram for the beam. Ignore its weight. b. Calculate the reaction forces at A and B (units: N) c. Determine the normal and shear forces (units: N) and the internal moment (units: N m) along the cut. Problem 5 A beam with a length of 1.2 m and negligible weight is firmly attached as a fixed support on the left end. A mass of 4 kg, attached at x=0.6 m, pulls down to create force W. A cable pulls up with force P of 100N at x=0.2m, and a force R of 30N is applied to the end of the beam. a. Draw a FBD of the beam. b. Calculate the values of the reaction forces (units: N) and moment (units: Nm) generated at the fixed support. c. Calculate the internal forces and moment at the point A. 0.6 m 0.6 m 0.4 m 0.2 m 3 m 1 m 40° 900 N
Tae Tae T EM a TC 3 90061 TaosCho TsinCud Tf Tnc E Fx 900 t TO CYO t Fax O E it y Ts i n no T t Fay o 1 SE scyoj s.na 9871.8910 FAX 900 Tco Cyo 8 462.307N Fay Tsin Cao T 16217.420 N Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com
Af Agt py B m I wosncss.mg oocas apg.pzwAx o Caller Efx o Bxtcwicss.is 0 Bx looocoscss.is ooooo n sey o ay By esmc s.nl D By btcsincss.is ay By too too since.is 933.332 566.667 N Normal Forces Goo o i n Nx shear Fees 566ait N Ny Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com
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R 30N 70 C a A MS ooh oom o om Tx Pcos 70 R cos oo o Fx 0 Tx Pcos 7 offresco 100 cacao t 3 cases 49.202N EFy 0 Ty Psin 70 Rin Ga Wg 0 Ty Rsin oo Ps n 7 tug 2 788 N M Mtp sin 7070.2 Wg o 6 Rs in Coo 1 2 o M Wgco 6 trs info 1 2 ns.nczolfo.z m 3 5.903 Nm Ay Rance 25 981N A x R co Ca is N Mint Rsn 18.70.4 co 392 Nm Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com
Problem 6 While relaxing in a hammock on vacation, Jane starts to think about biomechanics. If Jane weighs 55kg, determine the tension on the two support ropes as shown to the right (units: N). Angle A is equal to 45° and angle B is equal to 35°. Your F.B.D. will be of Jane connected to the tree by cable connections (i.e. the ropes). Ignore the weight of the hammock and ropes. Problem 7 You are performing a uniaxial compression test on a specimen containing a bone and a solid metal graft. The bone has two segments, a spongy, trabecular segment and a hollow, cortical segment (as shown at right). The outer section of the hollow bone and the entire cross section of the spongy bone can be treated as solid materials. Ignore the weight of the specimen in your calculations. The hollow bone has an inner radius of 0.5 cm. The graft and the bone have an outer radius of 1 cm. Before loading, each segment is 2 cm long. Suppose the hollow bone has a Young’s modulus (E) of 10 GPa, and the spongy bone has a Young’s modulus of 1 GPa. You load the bone with a 55 N force. After loading, the entire specimen’s length is 5.94 cm. a. Calculate the normal stresses in the hollow bone, spongy bone, and graft (units: MPa). b. Calculate the change in length in each of the three segments (units: mm). c. Calculate the Young’s modulus of the graft (units: MPa) Problem 8 After the compression test in problem 7, Rebecca unloads the specimen and it returns to its original dimensions. Now she applies an 80 N m torque to the top of the specimen, which is firmly fixed to the surface at its base. a. Calculate the shear stress at the inner and outer radius of the hollow bone (units: N/cm2). b. With the same torque in place, calculate the maximum shear stress in the solid metal graft (units: N/cm2). F = 55 N
44 w 55 9.8 1 5 I 539N NW Fx A cos us B cos 635 0 A cos 451 13801135 A II Fy 0 A sin 45 Bsin 35 W O Asinasst Bs no 5 539 Bws9 gg t Bsiness 539 B 1392728 1 539 13 387.0180225 A 429179 0135 448.3433 Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com
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a Sponge I man random man 0.175 MPa 0.2334 MPa 0.175 Mpa Extenghorneld ILI E Ate 41 6 5.94 Din Des 0.175 MPa 20mm e o 23340hPa 20mm 1 Gpa 10 Gpa 3.501 10 3mm 4 669 10 mm 0.596 mm c E Y 0 175120 5.8726hPa Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com
8 a T Ii 8000 0.5 2716.28 Man J r 4 5 1211 o s To Ig 8 2262 5432.5682 1.47262 Jo E 1 1.571 To 89 1,1 5092 958 Nkm2 Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com