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 = 3.68 104 645 153 = 36.3 N/mm = qb,65 qb,18 = 3.68 104 1290 153 = 72.6 N/mm qb,78 = 0 (from symmetry) (from symmetry) qb,76 = 3.68 104 645 (153) = 36.3 N/mm = qb,32 (from symmetry) qb,63 = 36.3 + 36.3  3.68 104 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 midpoint 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.
 Winter '08
 VALDEVIT

Click to edit the document details