152 Mechanics of Materials MA=- f60+7.5x1024-x)xdx 42 、 103 16 (30+7.5x)(16-8x+x2)xdx 103 16 (480x-240x2+30x3+120x2-60x3+7.5x4)dx 103 (480x-120x2-30x3+7.5x4)dx 16 103[480x2120x330x4,7.5x374 -162-3-4+50 103 16240×16-40×64-30×64+2.5×1024] =-120×103Nm The required moment at A is thus 120kNm in the opposite direction to that shown in Fig.6.8. Problems 6.1 (A/B).A straight beam ABCD is rigidly built-in at A and D and carries point loads of 5kN at B and C. AB=BC CD=1.8m If the second moment of area of the section is 7 x 10-m+and Young's modulus is 210GN/m2,calculate: (a)the end moments; (b)the central deflection of the beam. [U.Birm.][-6kNm;4.13mm.] 6.2 (A/B).A beam of uniform section with rigidly fixed ends which are at the same level has an effective span of 10m.It carries loads of 30kN and 50kN at 3m and 6m respectively from the left-hand end.Find the vertical reactions and the fixing moments at each end of the beam.Determine the bending moments at the two points of loading and sketch,approximately to scale,the B.M.diagram for the beam. [41.12,38.88kN:-92,-90.9,31.26,64.62kNm] 6.3 (A/B).A beam of uniform section and of 7m span is "fixed"horizontally at the same level at each end.It carries a concentrated load of 100 kN at 4 m from the left-hand end.Neglecting the weight of the beam and working from first principles,find the position and magnitude of the maximum deflection if E=210GN/m2 and 【=190×106m4. [3.73 from Lh.end;4.28 mm.] 6.4(A/B).A uniform beam,built-in at each end,is divided into four equal parts and has equal point loads,each W,placed at the centre of each portion.Find the deflection at the centre of this beam and prove that it equals the deflection at the centre of the same beam when carrying an equal total load uniformly distributed along the entire length. [U.c.L1] 「WL37 96E1152 Mechanics of Materials .. MA = - / (30 + 7.5x)103 (4 - x)~ x dx 42 0 4 103 - - - J (30 + 7.5~)(16 - 8x + x2)x dx 16 0 4 103 = - - 1 (480~ - 2402 + 30x3 + 1202 - 60x3 + 7.5~~) dx 16 0 4 103 - - - - 1 (480~ - 120x2 - 30x3 + 7.5x4)dx 16 0 103 = --[240~16-40~64-30~64+2.5~1024] 16 = -120x103Nm The required moment at A is thus 120 kN m in the opposite direction to that shown in Fig. 6.8. Problems 6.1 (A/B). A straight beam ABCD is rigidly built-in at A and D and carries point loads of 5 kN at B and C. AB = BC =CD = 1.8m If the second moment of area of the section is 7 x 10-6m4 and Young’s modulus is 210GN/mZ, calculate: (a) the end moments; (b) the central deflection of the beam. [U.Birm.l[-6kNm; 4.13mm.l 6.2 (A/B). A beam of uniform section with rigidly fixed ends which are at the same level has an effective span of 10 m. It carries loads of 30 kN and 50 kN at 3 m and 6 m respectively from the left-hand end. Find the vertical reactions and the fixing moments at each end of the beam. Determine the bending moments at the two points of loading and sketch, approximately to scale, the B.M. diagram for the beam. c41.12, 38.88kN; -92, -90.9, 31.26, 64.62kNm.l 6.3 (A/B). A beam of uniform section and of 7 m span is “fixed” horizontally at the same level at each end. It carries a concentrated load of 100 kN at 4m from the left-hand end. Neglecting the weight of the beam and working from first principles, find the position and magnitude of the maximum deflection if E = 210GN/m2 and I = 1% x m4. C3.73 from 1.h. end; 4.28mm.l 6.4 (A/B). A uniform beam, built-in at each end, is divided into four equal parts and has equal point loads, each W, placed at the centre of each portion. Find the deflection at the centre of this beam and prove that it equals the deflection at the centre of the same beam when carrying an equal total load uniformly distributed along the entire length. [U.C.L.I.] [--.I WL’ 96~1