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Problem 5 (6 marks) For the beam and loading shown, a) Draw the shear and bending-moment diagrams. b) Determine the maximum absolute value of the shear and bending moment. c) Determine the minimum required depth h of the beam. Knowing that the material used has an ultimate normal and shear stresses, use the given factor of safety in each loading case in the following table. P A LLL Dimensions, material properties and load values are presented in the following table for each loading case. Group No. Factor of Safety 2.7 1.2 3.3 Ultimate Strength LL Lb el ot mm mm mm mm kN kN/m MPa MPa 1.5 1.5 550 330 1.5 455 275 0.9 0.8 0.8 40 450 240 1.3 470 280 0.7 315 185 1.4 1.3 260 0.8 1 0.9 570 330 2 2.25 2.25 390 200 18 171. 880 189510 510300 165 3.2 2.8 3.1

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Answer 1

Q. Given: from the material properties table given analyse group Group G7 L .8 in mm 22 L₂ 6 1 .9 25 KN kN/m MPa Mpa P W Jult Tult 14 7 570 330 factor of Safety 3.2. р Shear Force Diagam (Drawing from left to right) C L2 D L3 B 5 6N (WL + p) = 14.0056 KN

Bending Moment Diagram. 0<x< 8 (Ato c) M = - COX X X = -W? = -3.53? Nmm f<x< 1.8 ( to D) M - XL, X X-4/2) = -W 4,3 +6012 - - 7X-83 + 7x.82 2. M = -5.6x + 2.24 Nam - Hex < 2.7 to B M= -X(P+w4) + PCL4+12)+W42 = -x (14000 + 7%. B) + 14000 (18) + 3.58.82 = -xX14005.6 + 25 200+ 2.24 Nmm. = -14005.6x + 25202.24 e = -2.24 Nmm. = -7.84 Nmm B = -12672.88 Nmm.

Bending Moment Diagram, - 4 C 2 0 13 B -2. 241 -7.84 - 26 12 88 0 Maximum Value of Bending moment and shear Lonice, from the Shear force and Bending Moment Diagram Max Shear force occurs at point B Fmax = -14005.6 N. Max. Bending Moment Occuns at point B Mmax = -126/2.88 Nmm.

Minimum value of 'n' Max stressed/ Cultural element will be at point B. Bending Stress at pocnt B OB= - Mah - 2x x bx h3 bh² GMB N/mm² Shear fotce at Point B will exent directshear Stress TB = N 14005.6 bh. subsituting values of MB, b ir 1009 Te = 166.7 h2h so Principal stresses at pont B are The = 500. 5045° 166-72 72 2 - 2

Now Suit = 570 N/mm2 Tsult = 330 N/mm². F.S= 3.2 Allowable stresses - Salu = 570 = 178.12 3.2 Call = 330 = 103.12 Loing Max. Normal stress failure. - 178.12 = sous 1 5o4.5² 186.7.12 h2 h² taking h²=2 (178.12.3 - S04.5) = 504.52 + 186.7?36 178-123234.504.= 2x178:12X804.5X = so4.5°+186-7X 178.12?x2-186.72% - QX178.12.X 504.5% = 0 176.12 x _ 6.768) =0 so x²= 6.762 x=6.76 and y=h2 : h= 2.6 mm. - Answers

Now using Max.Shear theory Tall = (01-02) 163:12 = soy.524186.742 again taking hex (103:12x) = sou.s?+ 186:7-X 103.12?x?-18675 -504.5+=0 X-3.27 x - 23.93 - 0 X = 6.79 as h²=x h = 2.605 mm Auswer (2) So A nower on Both accounts of failure h = 2.6 mm. - Final Answer.