Assignment6_Solutions

# Assignment6_Solutions - S.23.6 Referring to Fig P.23.6 the...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: S.23.6 Referring to Fig. P.23.6, the horizontal x axis is an axis of symmetry so that Ixy = 0. Further, Sx = 0 so that, from Eq. (20.6) qb = - in which Ixx = 4 1290 1532 + 4 645 1532 = 181.2 106 mm4 Eq. (i) then becomes 66 750 qb = - 181.2 106 n Sy Ixx n Br yr r=1 (i) Br yr = -3.68 10 r=1 -4 n Br yr r=1 Now, `cutting' Cell I in the wall 45 and Cell II in the wall 12 qb,45 = 0 = qb,12 qb,43 = -3.68 10-4 645 153 = -36.3 N/mm = qb,65 qb,18 = -3.68 10-4 1290 153 = -72.6 N/mm qb,78 = 0 (from symmetry) (from symmetry) qb,76 = -3.68 10-4 645 (-153) = 36.3 N/mm = qb,32 (from symmetry) qb,63 = 36.3 + 36.3 - 3.68 10-4 1290 (-153) = 145.2 N/mm The shear load is applied through the shear centre of the section so that the rate of twist of the section, d/dz, is zero and Eq. (23.10) for Cell I simplifies to 0= 1 [qs,0,I (34 + 45 + 56 + 63 ) 2AI GREF - qs,0,II 63 + qb,63 63 + qb,34 34 + qb,56 56 ] and for Cell II 0= 1 [-qs,0,I 63 + qs,0,II (12 + 23 + 36 + 67 + 78 + 81 ) + qb,81 81 2AII GREF + qb,23 23 + qb,36 36 + qb,67 67 ] (iii) (ii) Solutions to Chapter 23 Problems 281 in which GREF = 24 200 N/mm2 . Then, from Eq. (23.9) t34 = t56 = t36 = t81 20 700 0.915 = 0.783 mm 24 200 24 800 1.220 = 1.250 mm = t45 = 24 200 Thus 34 = 56 = 380/0.783 = 485.3 12 = 23 = 67 = 78 = 356/0.915 = 389.1 36 = 81 = 306/1.250 = 244.8 45 = 610/1.250 = 488.0 Eq. (ii) then becomes 1703.4qs,0,I - 244.8qs,0,II + 70 777.7 = 0 or qs,0,I - 0.144qs,0,II + 41.55 = 0 and Eq. (iii) becomes -244.8qs,0,I + 2046qs,0,II - 46 021.1 = 0 or qs,0,I - 8.358qs,0,II + 188.0 = 0 Subtracting Eq. (v) from (iv) gives qs,0,II = 17.8 N/mm Then, from Eq. (v) qs,0,I = -39.2 N/mm The resulting shear flows are then q12 = q78 = 17.8 N/mm q32 = q76 = 36.3 - 17.8 = 18.5 N/mm q54 = 39.2 N/mm q63 = 145.2 - 17.8 - 39.2 = 88.2 N/mm q43 = q65 = 39.2 - 36.3 = 2.9 N/mm q81 = 72.6 + 17.8 = 90.4 N/mm Now taking moments about the mid-point of the web 63 66 750xS = -2 q76 356 153 + 2 q78 356 153 + q81 306 712 - 2 q43 380 153 - q54 2(51 500 + 153 380) (see Eq. (20.10)) from which xS = 160.1 mm (v) (iv) S.23.4 In this problem the cells are not connected consecutively so that Eq. (23.6) does not apply. Therefore, from Eq. (23.5) for Cell I 1 d = [qI (12U + 23 + 34U + 41 ) - qII 34U - qIII (23 + 41 )] dz 2AI G (i) Solutions to Chapter 23 Problems 277 For Cell II 1 d = [-qI 34U + qII (34U + 34L ) - qIII 34L ] dz 2AII G For Cell III 1 d = [-qI (23 + 41 ) - qII 34L + qIII (14 + 43L + 32 + 21L )] dz 2AIII G In Eqs (i)(iii) 12U = 1084/1.220 = 888.5 12L = 2160/1.625 = 1329.2 (iii) (ii) 14 = 23 = 127/0.915 = 138.8 34U = 34L = 797/0.915 = 871.0 Substituting these values in Eqs (i)(iii) d 1 = (2037.1qI - 871.0qII - 277.6qIII ) dz 2 108 400G d 1 = (-871.0qI + 1742.0qII - 871.0qIII ) dz 2 202 500G 1 d = (-277.6qI - 871.0qII + 2477.8qIII ) dz 2 528 000G Also, from Eq. (23.4) 565 000 103 = 2(108 400qI + 202 500qII + 528 000qIII ) Equating Eqs (iv) and (v) qI - 0.720qII + 0.075qIII = 0 Equating Eqs (iv) and (vi) qI - 0.331qII - 0.375qIII = 0 From Eq. (vii) qI + 1.868qII + 4.871qIII = 260.61 Now subtracting Eq. (ix) from (viii) qII - 1.157qIII = 0 Subtracting Eq. (x) from (viii) qII + 1.853qIII = 100.70 Finally, subtracting Eq. (xii) from (xi) qIII = 33.5 N/mm (xii) (xi) (x) (ix) (viii) (vii) (iv) (v) (vi) 278 Solutions Manual Then, from Eq. (xi) qII = 38.8 N/mm and from Eq. (ix) qI = 25.4 N/mm Thus q12U = 25.4 N/mm q21L = 33.5 N/mm q14 = q32 = 33.5 - 25.4 = 8.1 N/mm q43U = 38.8 - 25.4 = 13.4 N/mm q34L = 38.8 - 33.5 = 5.3 N/mm ...
View Full Document

## This homework help was uploaded on 04/07/2008 for the course MAE 157 taught by Professor Valdevit during the Winter '08 term at UC Irvine.

Ask a homework question - tutors are online