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MM222Strength of MaterialsLecture 07Spring 2015Hafiz Kabeer RazaResearch AssociateFaculty of Materials Science and Engineering, GIK InstituteContact: Office G13, Faculty Lobby raza@giki.edu.pk, hkabeerraza@gmail.com, 03344025392Overview of the type of stressesType of stressExperienced byCross-section StressNormalPlate/ sheet/ member/ supportCross-section Normal to the applied forceP/AWhere A= tw for bomm or d2/4 for rodShearBoltCross-section being shearedP/AWhere A= d2/4Bearing Plate/ sheet/ member / supportArea of member in-contact with the boltP/AWhere A=tdt= plate thicknessd= dia of boltMaximum Normal stressPlate/ sheet/ member / supportCross-section Normal to the applied force And Near any hole (or inhomogeneity)P/AWhere A = t(w-d)

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsSteps in problem solvingNote down the stresses required [the portions of interest]Determine the forces at supports [only where required]Determine the internal forces in members [only for which the stress is asked]Note down the dimensions of the members, joints, bolts, pins, supports etc [of interest]For each stress required, carefully note down the value of forceCalculate the stresses.

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsSample Problem 1.1Stresses required:A (single-shear)C (double-shear)ABC (maximum normal)B(bond) (double-shear)b(C) (bearing in link)

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsSample Problem 1.1Find the reactions at supportsFree body DiagramMD = 0-FAC 10 in + 500 lb 15 in = 0FAC = 750 lbDimensions (inches)Upper ABC: t=3/8Lower ABC: t=1/4dA = 3/8dC = Max-normal area = t(w-dA)Bearing area = dt = 0.250.25ForcesA (single-shear) 750 lbC (double-shear) 375 lbABC (max-normal) 750 lbB(bond) (double-shear) 375 lbb(C) (bearing in link) 375 lb

StressesA (single-shear) = 6794 psiC (double-shear) = 7643 psiABC (max-normal) = 2285 psiB(bond) (double-shear) = 171.4 psib(C) (bearing in link) = 6000 psi

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsProject workProblems 1.1 and 1.2Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsQuizzesStarts from this week

Academic Week #Quiz #Date (Wednesday)3125-02-20155211-03-20157325-03-20159416-04-201511530-04-201513614-05-201515728-05-2015Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsSample Problem 1.2What to find?Diameter (d) of bolt = diameter of hole = inner dia of the endDimension b = outer dia of endDimension h = regular width

Spring 2015By Hafiz Kabeer RazaMM222 Strength of MaterialsSample Problem 1.2Diameter (d) of bolt = diameter of hole = inner dia of the endAllowable shear stress will be considered (100 MPa)Dimension b = outer dia of endAllowable normal stress will be taken (175 MPa)Dimension h = regular widthAllowable normal stress will be taken (175 MPa)

For d = P/Shearing area Area = 3.14/4*d2d = 27.6 mm 28 mmCross check the allowable bearing stressb = 214 MPa < allowable bearing stress (Safe)For b = P/maximum normal area Area = t*(b-d)b = 62.3 mm 63 mmFor h = P/Normal areaArea = h*th = 34.3 mm 35 mmSpring 2015By Hafiz Kabeer RazaMM222 Strength of Materials