# CA1 - MEM 230 Mechanics of Materials Spring Term 2008/09...

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Unformatted text preview: MEM 230 Mechanics of Materials Spring Term 2008/09 April 1, 2009 Class Assignment # 1 Name: ID# In the hanger shown the upper portion of link ABC is 0.375-in. thick and the lower portions are 0.25-in. thick. Epoxy resin is used to bond the upper and lower portions together at B. The pin at A is 0.375-in.-diameter while a 0.25-in.—diameter pin is used at C‘. Determine: The shearing stress in pin A. The shearing stress in pin C. The largest normal stress in link ABC. The average shearing stress on the bonded surfaces at B The bearing stress in the link at C. sun-99‘s 5 in. 1.75 in_-—~ T in. y "#00 ii: MEM 230 Mechanics of Materials Spring Term 2008!09 ' _Apﬁll,2009 Class Assignment # 1 SAMPLE PROBLEM 1.1 in the hanger shown the upper portion of link ABC is g in. thick and the lower portions are each i in. thick. Epoxy resin is used to bond the up 1' and lowsr portions together at B. The pin a: A is of %-ln. diameter while. a rim—diaman- pin is used at C'. Determine (a) the shearing stress in pin A, (b) the shearing siresn in pin (2‘. (c) ll‘lii.‘ largest normal stress in link MIC. (of) this average shear ing stress on ll'ic handed surfaces at B. (a) the bearing stress in the link :11 C. SOLUTION Free Body: Entire Hanger. Since the link ABC is n two-force. member, the reaction at A is vertical; the. reaction at D is represented by its camponents D, and By. We write +’i 2M5 2 U: {5001b}{15in.) — FA¢{1(}in.}:L) FM: : +750 lb FM- n 7501b mm)” a. Shearing Stress in Pin A. Since this ﬁ-inndiametcr pin is in single shear. we write Fm 758 111 = m TA mun psi 4 'r ==-—~ " A ﬁn(0.375in.}7 b. Shearing 511155 in Pin C. Since this ﬁlm-diameter pin is in (Enable shear. we write .- 23!} iii I F ~ 37 l . #25 “2W5 1 7(ﬁ764ilpsi‘1 I A gal-[0.25 m.) j PM —in. (linihettr _ A ‘ inmydiamlw ii Fir m“, c.‘ largest Normal Stress in Link ABC. The largest stress I5 tuund ' _ when: the area is smallest; this. occurs at the cross section at A where the ﬁ-in. It "2111. Far 75”“1 hole is located. We have :1" 5 . 1"”“1' 3 FM.— 750 in 7501b g4 - - ——--—- Uﬂ «T- 22'9i'ipi: 4 A”. * E3 in.)(1.25 in. A 635%.) L 0.3231n1 1!, Average Shearing Stress at B. We note that handing exists on bOIh sides of the upper portion of the link and that the shear force on each side is F, = (7501b)! 2 w: 3751]). The average: shearing stress on each surface is thus 375 it: (1.25 i'n‘)(l.75 in.) \ :14". diameter i i F. a: . i} gr“. - .175 ii. F. 7.3:.(Tl: 7R:!7[‘4133~‘i “ e. Bearing Stress in Lin ai C. For each portion of the link. Fl == 375 1b and the nominal bearing area is (0.25 in.}(0.25 in} = 0.06251'113. F. 375]]: = q a —— (>000 .' 4 0” A 0.0625 in2 ‘r" p“ - i 4 -I.II.. diam: 1.9.5 ...
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## This note was uploaded on 09/23/2009 for the course MEM 230 taught by Professor Awerbuch during the Spring '08 term at Drexel.

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CA1 - MEM 230 Mechanics of Materials Spring Term 2008/09...

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