April 6, 28 UNIVERSITY O RHODE ISAND Deprtment of Electricl, Computer nd Biomedicl Engineering BME 27 Introduction to Biomechnics Spring 28 Homework 8 Prolem 14.6 in the textook. In ddition to prts -e, t section - (x = 2 m) compute f) the mximum norml (flexurl) stress σ, nd t section - (x = 3 m) compute g) the mxmum sher stress τ nd flexurl stress σ. Answers f) σ = 3 kp g) τ = 7.5 kp, σ = 15 kp Prolem 2 The uniform rectngulr simpl-supported em of length, width, nd height h is hinged t point A nd supported n idel roller t B. A force is pplied distnce d from the left end. Section - is locted etween x d nd section - etween d x. Assuming the weight of the em is negligile, clculte: A. the rection forces on the em t A nd B; B. the internl sher force V nd ending moment M t section -; C. the internl sher force V nd ending moment M t section -. D. Sketch the sher force V nd the ending moment M versus x, long the length of the em. Be sure to check our units. Confirm tht our sketches stisf the eqution V = dm/dx. E. Given the dimensions nd vlues: = 6 m d = 4 m = 75 N section - t 3 m = 1 cm h = 2 cm section - t 5 m i. Compute the mximum flexurl stress σ mx, mximum sher stress τ mx, nd verge sher stress τ vg t sections - nd -. ii. Sketch the flexurl nd sher stresses t sections - nd - on the profile. A B h d - 1 -
Prolem 2 continued BME 27 - Homework 8 April 6, 28 Answers A. R A = (1 d/), R B = (d/) B. V = R A = (1 d/), M = xv = x (1 d/) C. V = (d/), M = d (1 x/) D. Sketches: (d/) V, N (1 d/) d M, N-m d(1 d/) d E. section - t = : σ mx = 112.5 kp; τ mx = 1.875 kp, τ vg = 1.25 kp section - t = : σ mx = 75 kp; τ mx = 3.75 kp, τ vg = 2.5 kp Sketches for section -: σ mx = 75 kp τ mx = 3.75 kp σ τ σ mx = 75 kp τ = Note tht with pure ending lod, the stresses σ nd τ re smmetric out =, the neutrl xis. The sher stress mgnitude is mximum t = (in this cse, the neutrl xis), nd zero t = ±h/2 (the upper nd lower surfces of the em). - 2 -
BME 27 - Homework 8 April 6, 28 Prolem 3 The uniform rectngulr em of length is hinged t point A, supported n idel roller t B, nd hs two forces pplied distnce d from either end. Section - is locted etween x d, section - is etween d x d, nd section c-c etween d x. Assuming the weight of the em is negligile, clculte: A. the rection forces on the em t A nd B; B. the internl sher force V nd ending moment M t section -; C. the internl sher force V nd ending moment M t section -. D. the internl sher force V c nd ending moment M c t section c-c. E. Sketch the sher force V nd the ending moment M versus x, long the length of the em. el the xes ppropritel, nd confirm tht the sketches stisf V = dm/dx.. Given the dimensions nd vlues: = 4 m d = 1 m = 1 N section - t.5 m = 1 cm h = 2 cm section - t 2. m section c-c t 3.5 m i. Compute the mximum flexurl stress σ mx, mximum sher stress τ mx, nd the verge sher stress τ vg t sections -, -, nd c-c. ii. Sketch the flexurl nd sher stresses t sections -, -, nd c-c on the em profile. A c B h c d d This is sometimes clled xle loding ecuse it cn represent the lods on vehicle xle (for exmple, the forces could e the vehicle s weight nd the rections t A nd B could e the wheels). - 3 -
Prolem 3 continued Answers A. R A = R B = B. V =, M = x C. V =, M = d D. V c =, M c = ( x) E. Sketches: BME 27 - Homework 8 April 6, 28 V, N d d M, N-m d d d. section -: t =, σ mx = 75 kp; t =, σ mx = 75 kp; τ mx = 7.5 kp; τ vg = 5 kp section -: t = : σ mx = 15 kp; t =, σ mx = 15 kp; τ mx = kp; τ vg = kp section c-c: t = : σ mx = 75 kp; t =, σ mx = 75 kp; τ mx = 7.5 kp; τ vg = 5 kp Sketches t section -: σ mx = 75 kp τ mx = 7.5 kp σ τ σ mx = 75 kp τ = - 4 -
BME 27 - Homework 8 April 6, 28 Prolem 4 The uniform rectngulr cntilever em of length is secured to the wll t point A, nd supports the force t the free end. Section - is locted etween x. The weight of the em is negligile. A. Clculte the internl sher force V nd ending moment M t section -. B. Sketch the sher force V nd the ending moment M long the length of the em. Verif tht V = dm/dx. C. Given the dimensions nd vlues: = 3 m d = 1 m = 1 N = 5 cm h = 2 cm section - t.5 m i. Compute the mximum flexurl stress σ mx, mximum sher stress τ mx, nd verge sher stress τ vg t section -. ii. Sketch the flexurl nd sher stresses t section - on the profile. D. Compute the principl stresses nd mximum sher stress t x =.5 m nd =, =, nd =. How do these compre to our nswers in prt C? A x h - 5 -
Prolem 4 continued Answers A. V =, M = (x ) B. Sketches: BME 27 - Homework 8 April 6, 28 V, N M, N-m C. t = : σ mx = 75 kp; t = : τ mx = 15 kp, τ vg = 1 kp Sketches: σ mx = 75 kp τ mx = 15 kp σ τ σ mx = 75 kp τ = D. t x =.5 m: σ 1 σ 2 τ mx 75 kp kp 375 kp 15 kp -15 kp 15 kp kp -75 kp 375 kp - 6 -
BME 27 - Homework 8 April 6, 28 Prolem 5 The rectngulr cntilever em of length is secured to the wll t point A, nd supports the force t the loction x = d, s shown. Section - is locted etween x d ; section - is etween d x. The weight of the em is negligile. A. Clculte: i. the internl sher force V nd ending moment M t section -; ii. the internl sher force V nd ending moment M t section -. iii. Sketch the sher force V nd the ending moment M long the length of the em, nd verif tht V = dm/dx. iv. Given the dimensions nd vlues: = 3 m d = 1 m = 1 N = 5 cm h = 2 cm section - t.5 m 1. Compute the mximum flexurl stress σ mx, the mximum sher stress τ mx, nd verge sher stress τ vg t sections - nd -. 2. Sketch the flexurl nd sher stresses t section -. B. Repet the ove if the em dimensions re swpped to = 2 cm nd h = 5 cm. C. How do the mximum stresses chnge with the em dimensions? If the ending lod grows lrger, which em will fil first, the one in prt A or prt B? Wh? A d x h - 7 -
Prolem 5 continued Answers A. i. V =, M = (x d) ii. V =, M = iii. Sketches: BME 27 - Homework 8 April 6, 28 V, N d M, N-m d d iv. section -: t =, σ mx = 15 kp; τ mx = 15 kp; τ vg = 1 kp section -: for n, σ mx = kp; τ mx = kp; τ vg = kp Sketches t section -: σ mx = 15 kp τ mx = 15 kp σ τ σ mx = 15 kp τ = B. i-iii These nswers re unchnged. (Wh?) iv. section -: t =, σ mx = 6 kp; τ mx nd τ vg unchnged. Except for the mgnitude of σ mx, the sketches re unchnged. C. The second em (shorter nd wider, in prt B) will fil first. - 8 -
BME 27 - Homework 8 April 6, 28 Prolem 6 The rectngulr cntilever em of length supports forces 1 nd 2 s shown. Assuming the weight of the em is negligile: A. clculte nd sketch the internl sher force V nd the ending moment M long the length of the em, nd verif tht V = dm/dx. B. Given the dimensions nd vlues: = 4 m d = 1 m 1 = 1 N section - t.5 m = 1 cm h = 1 cm 2 = 2 N section - t 3. m i. Compute the mximum flexurl stress σ mx, the mximum sher stress τ mx, nd the verge sher stress τ vg t sections - nd -. ii. Sketch the flexurl nd sher stresses t section -, nd show the neutrl xis. C. Using the sme dt given ove, consider mteril element t =, where the flexurl stress is lrgest nd the sher stress is zero. Determine the principl stresses σ 1 nd σ 2 nd the orienttion of the principl plne. Sketch the principl stresses on properl oriented mteril element. D. On the sme em, compute the principl stresses nd their orienttions on mteril element locted t =. How do these differ from those ove? E. At = compute the mximum sher stress τ mx, the ssocited norml stresses, nd the orienttion of mximum sher stress plne. Sketch these on properl oriented mteril element. d 1 2 x h - 9 -
BME 27 - Homework 8 April 6, 28 Prolem 6 continued Answers A. Sketches: V, N 2 1 2 d M, N-m ( d) 2 d 1 2 d B. i. section - t = : σ mx = 4.5 MP, τ mx = 45 kp; τ vg = 3 kp section - t = : σ mx = 1.2 MP, τ mx = 3 kp; τ vg = 2 kp ii. Sketches t section - (x = 3 m): σ mx = 1.2 MP τ mx = 3 kp NA τ σ mx = 1.2 MP τ = C. section -: σ 1 = 4.5 MP, σ 2 = MP, θ P = section -: σ 1 = 1.2 MP, σ 2 = MP, θ P = D. section -: σ 1 = MP, σ 2 = 4.5 MP, θ P = 9 section -: σ 1 = MP, σ 2 = 1.2 MP, θ P = 9 E. section -: τ mx = 45 kp t θ S = 9 section -: τ mx = 3 kp t θ S = 9-1 -
BME 27 - Homework 8 April 6, 28 Prolem 7 The sme rectngulr cntilever em from Prolem 4 supports the sme 1 N lod t the free end. This prolem will exmine how the flexurl stresses nd mximum sher stress chnge in the x direction, long the length of the em. Compute nd plot the mximum flexurl stress σ mx (t = ) nd mximum sher stress τ mx (t = ) t.5 m intervls long the x xis, strting t x =.5 m nd ending t x = 2.5 m. Does the neutrl xis st t constnt loction? Does either stress chnge long the length of the em? If so, wh? If not, wh not? = 3 m = 1 N x 5 cm 2 cm - 11 -
BME 27 - Homework 8 April 6, 28 Prolem 8 The circulr cntilever em of dimeter d o supports force t the end of its length. Section - is locted etween the ends of the em, t x. Assuming the weight of the em is negligile: A. Clculte the internl sher force V nd ending moment M t section -. B. Sketch the sher force V nd the ending moment M long the length of the em. C. Given the dimensions nd vlues: = 1 m d o = 1 cm = 3 N section - t 9 cm i. Compute the mximum flexurl stress σ mx nd mximum sher stress τ mx t section -. ii. Sketch the flexurl nd sher stresses t section - nd show the neutrl xis. D. The solid rod must e replced with hollow tue where the inner rdius is one-hlf of the outer rdius. If σ mx must remin unchnged t section -, compute the outer rdius of the tue. E. The solid rod must e replced with squre em (i.e., h = ). If σ mx must gin remin unchnged, compute the height of the squre em.. The solid rod must e replced with rectngulr em with height h = 2. If σ mx must gin remin unchnged, compute the height of the rectngulr em. G. Assuming ech of the ove four ems re mde of the sme mteril, which geometr would provide the miniml weight em to support σ mx? x d o - 12 -
Prolem 8 continued Answers A. V =, M = ( x) B. Sketches: BME 27 - Homework 8 April 6, 28 V, N M, N-m C. i. t x =.9 m, = +r o : σ mx = 3.6 kp, τ mx = 5.9 kp ii. Sketches: +r o σ mx = 3.6 kp NA τ mx = 5.9 kp τ r o σ mx = 3.6 kp τ = D. r o = 5.11 cm for the tue with r i = r o /2. E. h = = 8.38 cm. h = 1.56 cm G. Ordered incresing volume: rectngle, hollow tue, squre, solid rod. So the rectngle is the miniml weight em needed to support σ mx. - 13 -
BME 27 - Homework 8 April 6, 28 Prolem 9 The simpl-supported em hs dimensions h = 4 cm, = 2 cm nd the ending moment digrm shown. 1 N 1 N 5 N A c c d d B h 2 m 3 m 5 m 6 m M, N-m 2 1 2 4 6 A. Crefull sketch the internl sher forces long the em s length. B. Compute the mximum ending stress nd mximum sher stress t x = 1., 2.5, 4., 5.5 m. C. Sketch the ending nd sher stresses on the cross-sectionl profile. Answers A: section sher force, N - V = 1 - V = 5 c-c V c = 5 d-d V d = 15 B (t = ): σ mx, MP τ mx, kp 1. 18.75-187.5 2.5 42.19-93.75 4. 37.5 93.75 5.5 14.6 281.25-14 -
BME 27 - Homework 8 April 6, 28 Prolem 1 The simpl-supported em hs the ending moment digrm shown. 1 N 5 N 1 N A B d i = 5 cm d o = 6 cm 1 m 3 m 5 m 6 m M, N-m 2 1 2 4 6 A. Crefull sketch the internl sher forces long the em s length. B. Compute the mximum ending stress nd mximum sher stress t x =.5, 2., 4., 5.5 m. C. Sketch the ending nd sher stresses on the cross-sectionl profile. D. Compute the principl stresses nd mximum sher stress t x = 2 m nd = +r o, =, nd = r o. Answers A: sect sher force, N - V = 125 - V = 25 c-c V c = 25 d-d V d = 125 B: σ mx, MP τ mx, kp (t = r o ) (t = ).5 5.69-287.79 2. 13.66-57.56 4. 13.66 57.56 5.5 5.69 287.79 D (t x = 2 m): stresses in MP σ 1 σ 2 τ mx +r o -13.66 6.83.578 -.578.578 r o 13.66 6.83-15 -
Prolem 11 The cntilever em hs the ending moment digrm shown. BME 27 - Homework 8 April 6, 28 11 11 11 11 11 11 11 11 11 11 A 4 N 3 N 2 N 1 N B r o = 4 cm 2 m 4 m 6 m 8 m M, N-m 2 4 6 8-2 -4 section -: M = 1x 4 section -: M = 6x 32 section c-c: M c = 3x 2 section d-d: M d = 1x 8 A. Crefull sketch the internl sher forces long the em s length. B. Compute the mximum ending stress nd mximum sher stress t x = 1., 3., 5., 7. m. C. Sketch the ending nd sher stresses on the cross-sectionl profile. Answers B: σ mx, MP τ mx, kp (t = +r o ) (t = ) 1. 5.97-26.53 3. 2.79-15.92 5..995-7.96 7..199-2.65-16 -
Prolem 12 The cntilever em hs the ending moment digrm shown. BME 27 - Homework 8 April 6, 28 11 11 11 11 11 11 11 11 11 11 A 3 m B r i = 3 cm t = 1 cm 6 m 8 m 5 M, N-m 4 3 2 1 2 4 6 8 A. Crefull sketch the internl sher forces long the em s length. B. Compute the mximum ending stress nd mximum sher stress t x = 1., 2., 5., 7. m. C. Wht is the vlue of the unknown force? Answers B: σ mx, MP τ mx, kp (t = r o ) (t = ) 1. 1.16 8.97 2..87 8.97 5..58. 7..29 8.97 C: = 1 N - 17 -
BME 27 - Homework 8 April 6, 28 Prolem 13 The circulr cntilever em of rdius r o nd length (just like the em in Prolem 8) supports force oriented t θ = 6 with the verticl. Since the loding force is not tngent or norml to the cross section, it genertes n xil stress, flexurl stress, nd sher stress within the em. (This comined loding prolem demonstrtes ll of the loding modes we hve studied, except torsion.) Section - is locted t x etween the ends of the em. Assuming the weight of the em is negligile: A. Given the dimensions nd vlues: = 1 m r o = 5 cm = 5 N section - t 9 cm i. Compute the xil stress, mximum flexurl stress, nd mximum sher stress t section -. ii. Sketch the xil, flexurl, nd sher stresses t section -. B. Sketch the net norml stress on section - dding the xil the flexurl stresses. Indicte the neutrl xis on the sketch. How is it different from the neutrl xis in prt C of Prolem 8? x θ = 6 r o - 18 -
Prolem 13 continued BME 27 - Homework 8 April 6, 28 Answers A. i. σ x = 5.5 kp; t = r o : σ mx = 25.5 kp, τ mx = 4.2 kp ii. Sketches t x =.9 m (section -): +r o +r o σ mx = 25.5 kp +r o τ mx = 4.24 kp r o σ x = 5.5 kp r o NA σ mx = 25.5 kp r o τ = τ B. t = r o : σ = 2 kp t = r o : σ = 31 kp Sketch of the net norml stress: +r o NA σ = 2 kp r o σ = 31 kp - 19 -
Prolem 14 The rectngulr em shown hs the dimensions nd vlues: BME 27 - Homework 8 April 6, 28 = 4 m d = 3 m 1 = 2 N = 5 cm h = 2 cm 2 = 2 N At the loctions x = 1, 2, 2.8, nd 3.5 m: A. compute the xil stress; B. compute the mximum flexurl stress nd mximum sher stress. C. Sketch the xil, flexurl, nd sher stresses. D. Sketch the net norml stress. Indicte the neutrl xis. Wht hppens to the neutrl xis from x = to 3.5 m? Is the result different from tht in Prolem 7? d 2 1 x h - 2 -
Prolem 14 continued Answers BME 27 - Homework 8 April 6, 28 A. σ x = 2 kp t ll x loctions B. t = : for x = 1. m: σ mx = 12 kp, τ mx = 3 kp for x = 3.5 m: M = so the flexurl stress σ mx =, nd V = so τ mx = C. Sketches t x = 1 m: σ mx = 12 kp τ mx = 3 kp σ X = 2 kp NA τ σ mx = 12 kp τ = D. Sketches: t x = 1. m t x = 3.5 m σ = 14 kp NA σ = 2 kp σ = 1 kp - 21 -