hw6-10sol - a) ) Hw Ho sow‘nws Stress—Strain Diagrams...

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Unformatted text preview: a) ) Hw Ho sow‘nws Stress—Strain Diagrams ’ Permanent Set The figure below plots four stress—strain curves for four different types of material. The specimens are loaded to two times the yield strain of the material. Rank these‘situations, from greatest to least, on the basis of the permanent set of each specimen. , n o M I 3. e . 61 '20 1110‘ — 7 a.» 77 e .30 E . 5/36” r, 6— I go- , :16 El 'ml 10 “*9 4 3m- ‘3 3 gum : i C- ;Ig—E 4 I e ’ fi 6 wk") 1 L1 6 :sz _ 2 I) '71" 1’ We 1 e 2130 Greatest 1 A 2 ’9 Or, the permanent set is the same at every point. Please carefully explain your reasoning. - . flu mudfia’L‘M-i . Fwy an 0'7— xw‘ & <11“, Lg E ‘ HIM «. ' ' L) ; (é!— C‘: M7, fax-WY _ 0101!? E i 01945) ' V1 . » _ C21 » eeowtflj’ _ eug‘y 6M“, Yfllfivl)’ ~ ' 3 E) 3 ‘5 q _ fl} _— -‘ Z — e » _ : _7 PA Etwwjl' Zia 0' 7 r 0’ 76 we r \ 5 i '; —‘- e ' l — -—\- 4 *5] eew=ae”’<§‘7£5¢ 4 re're I ' L - ~‘ 1 2 l _ 1 _ E. l C) EW#”4C \(oo. 59’~ 4e 43594556‘ m \__0.£.C:)=Le,i_ie_ EMMA 16' (6 ’5 r 1 46 v 4V , How sure are you of your ranking? (Circle one) _ Basically Guessed Sure Very Sure 1 ' 2 3 4 ' 5 6 7 ’ 8 9 10 26 cww ., ‘55 \O‘WO \w ._>’ a": '4— choow 4’ V =O,?‘3 5', '- 0:0 a; ’E "SM—‘4 W49- 0] (Lo *‘ (“WU unfwae, ‘ SmMRE MXHRE WUNE SOWME SmeE mmmmm 4;”; 0L» 4" W300 w WT'EWNQ but wwmw 6“ Y - l = 4"“ "33.5%“41 ‘ [6‘ ~90 M1 ' A “1“” 630%)" ‘ . ' i ~'6 '1’“ LW :5 ref - a‘ :mn'mgu; - A hz‘ 6"“ E ' mucous; e O‘Ooawofi; 0) AL: saw 0—) 1.0.003‘bq<{ 0.513 : moon‘s?“ : 0,00 I’M V 1 *C‘U'uJ ens-.1 Eda-1 ‘ '-V Lemfl =, 10~35M'°'°°3”°°W\ = °*°°“°“4 A; s e,ou u 0.00\\014(0.L:u\ = 0.000Lt‘3l95 UHF.) Q Ame (10+ M : 01.2» + 0.00010“, :“ : Moon.ch ___._..___._—_. .———-—-——-—-—-———-——-¢ a] VMM 91% L with GWEN: Mmm WA ‘31,, L0 = ’51“ 1’ = C; L ham \2 =°woo \L.,; k —. L53. +¢ =\.50'o\31'.\. uJuJLuLuLu SOUAH SOUAR SOUAR SOUAR Inn-unmnn 2° 5 42881 50 SHEETS EYE-EAS 42-382 100 SHEEFS EYE-EAS 42-389 200 SHEETS EYE~EAS 42-392 100 RECYCLED WHITE 23 2" f— A r 9 4, .9 T f r E.— xrin ' \500625 - L55. . —— = ~———-——-—-———— = 0.00003‘60 E +0 “~53! i . 5:3 Jik :‘qfilg'. eh:waka kimo- *«rfiful ' I ' 11f” Awmww. ’1. ' A’Vam (F - I“; " ‘ I C I: -- 7. '—---—-~ 1'. «Avid “1”” E «zoom ‘- GuT-ivl-M ' o'womu a) y: -eva m ~6,oooo%<so = 0.313 0 k4: ‘ kw f:ou k. = ’12» + Quooomoyufl s iooonb‘m 3—31. The shear stress—strain diagram for a steel alloy is shown in the figure. If a bolt having a diameter of 0.75 in. is made of this material and used in the double lap joint, determine the modulus of elasticity E and the force P required to cause the material to yield.Take v 7- 0.3. mm) , 60 The shear force developed on the shear planes of the bolt can be determined by considering the equilibrium of the FBD shown in Fig. a 7(rad) i>2F,,=0; V+V—P=0 v==g 0-00545 . From the shear stress—strain diagram, the yield stress is 1-, = 60 ksi.Thus, a V V P 2 z (0.75 ) V P = 53.01 kip = 53.0 kip . Ans. Ca From the shear stress—strain diagram, the shear modulus is _ 60ksi _ 3 . G — 0.00545 — 1101(10 )ksr Thus, the modulus of elasticity is E 0:55:75; 11.01003) = 2(1 + 0.3) E = 28.6003) ksi Ans. ...
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hw6-10sol - a) ) Hw Ho sow‘nws Stress—Strain Diagrams...

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