Question

Homework & Consider the three-point loaded beam below flexure P - loading pin a dollection (h=12") support roller & roller - P23 K6 6 8 = 21 The beam is loaded until a deflection = 0.2" is reached at the center. The rollers are steel RA= k = 0,25" ,E=30 Msi, te=42ks, RB = 0.5" The beam is made from A2318 Magnesium alloy (ty = 22ksi). E= 6.5 Mai Calculate is for gilding due to contact pressure for the aluminum beam at points AB and C -

it says a beam is loaded until a a deflection of 0.2 the rollers are steel Ra=Rc which is 0.25 and Rb= 0.5 modulus of elasticity is E=30 msi sigma y = 42 ksi the beam is A2318 magnesium alloy sigma y= 22ksi E=6.5 base =0.5 h=1 its asking to calculate factor of safety for yielding due to contact pressure for the beam at points A,B, and C don't know any other info that you will need

Answer 1

here factor of safety is calculated by finding out points on the beam where stress is more

which is calculated by finding loads at all given point diving the by areas of rollers as the forces on the
Page No. y (jertical) De Chronzatat 20 inch 6inch 6 ruch 16 8 = 0.2inch as This is the case of Central loaded beam so o deflection formulae in this case will be equal to id=PL -0 E = 6.5 Mego passed u8 EI Praha I = I bd3 12 here here as the Beam is loaded in transuare. veutical direction the Area moment will be calalated along x-x aris as bending is along x-x ans Put given values in Ixx = 150 = I like •ея" 0 1 xx = inch 0.2 = P(1213 24 u8x6.5x108x1 L = 12 inch 24 1504.6296 Pounds! IP =

Page No Now as we know from force balanco 15046296=2p, NOTE - (P = P, Because The given load pis Vymmetrical about Both hinge A 8 67 P, = 752.3148 - now we have load at Points PA = 752.3148 Pounds PB = 1504.6296 Pounds Pe= 752-3148 Pound at So we can All these Calulate Stress o points Relleus TA= = P Area 752.3148 ICRA RAC 0.25 inch Arear TR² TA = 752.3148 = 3831.5078 Powd Rina?? ZX10:25)

Page No. at will be same as of on because load and An rollers will be same JAROC = 3831.5078 Pounds (Pncul OR are ~ 3,831ksil OB = P = P X(R62 x (0.5) = 1504.6296. 7 (0.5)2 108 -1.915 ksid As we can see that stron is more at Point A & C so failure as Beam will ocam at Point A &c. Now we know FS - Factory of Safety = o tailun Gond yield Opermissible Tyield = 21ksi & Peen= =38316 final dhewe so IFs = 22 ki = 5.74 3.831 kj
beam is transferred to roller.