2 MOano1 363209 YTieRaVING MECH204P 2015/16 Answer any FIVE questions. All questions carry equal weight. ROUTE JANAETI I RO MOITANIMAX Note: A list of symbols, physical properties and useful equations are included in the last page of the examination paper. HADSHOSM 1. A simply supported horizontal beam of length L = 1.8 m is subjected to a concentrated downwards load F = 0.8 kN, applied at a distance L/4 from the left- hand support, and an additional concentrated force equal to F/2, applied at a distance L/3 from the right-hand support and directed upwards (see Figure 1.a). The beam is made from 3004-O aluminium, and has a hollow cross section as shown in Figure 1.b, where a = 3 cm and r = 1 cm. 2 Using the principle of superposition, sketch the shear force and bending moment diagrams, and estimate the maximum bending moment. [14 marks] CaWO! b. Calculate the maximum stress and the minimum radius of curvature for the beam. lensev aimsosos ontwoloo[6 Marks] NIT win (a) Figure 1. Schematic representation of the beam (a) and of the' cross section (b). RAVO MRUT HOBM-andtos\fLL Step 3 Super position of Bending Moment Between F, and F2 Mmax = Max of I Matcul or Mateys) In this case MMax - 0 5604 KN Xm (since this is higher than the amount moment due F. final Step The Maximum bending moment max in the beam is approximately 1 1max = 0 5604KN X m This moment is located at the Point of application F, DOMS\fLL R B X 1 . 8 = 0 8 x 0 45 + 0.y X 1 . 2 RB X 1 8 = 036 to 48. O 84KN XM RB = 0 84 = 0.467 KN 1 . 8 R B = 0. 467 KN 3. Reaction at A RA = 0 4 - RB = -0 067 KN RA s Negative which means "an upward reaction at A DOMS_ LL Part a Step 1 Determine Reaction forces at the support. F=08 KN acting downward 7/y = 0.45 F2=0.4 KN actiy upward ! / 3 = 0.6 1.Sum ofvertical forces RAt RR = F - 2 = RA + R B = 0.804 RAt RB - OUKN 2laking moment about Support A R B X L = F , x ( L / 4 ) + F 2 x (21/3 ) DOMS