Question

5 of 6 Learning Goal: To analyze two built-up members that have the same geometry but are fastened differently, determine the maximum applicable shear force on each cross section, and determine the adjustment in spacing between the weaker member's fasteners that would allow the member to support the equivalent maximum shear force of the stronger member The two cross sections shown below, (a) and (b), are subjected to a vertical shear force as shown. The members are fastened by nails that can support a load of 21.00 KN each and are spaced perpendicularly to the page in increments of s = 115 0 mm. The geometries of the cross sections are given by a = 250.0 mm, b = 40.00 mm c = 230.0 mm d = 170 mm, and e = 270 mm MITT b be (a) (b) Assume the cross sections are uniform along the entire lengths of the members.

Please find the maximum applicable shear force on member a and member b. Then the required spacing in the weaker member such that both cross sections can support the same maximum applicable force as defined by the stronger member.

Having trouble with the equations and I'm not really sure where I'm going wrong. Work would be appreciated. Thank you!!

Answer 1

(los Given data Load, v 21 KN S - 115 mm - 250mm o Mimon=9 c = 230mm d=170mm e = 270mm for cross section of member (a) EAY centroid , ý EA [2xh 2x bxсxx с ta ax cel Il 2 (bxc)+(axb) b{c+ 40 x 320ran] * [use xuo150497) to 2140 x 2 30)+(250X40) ý = = 162.5 mm

2 Moment of inertia , I = I x +Ady 2 qdxq I - 2 12 (exc) 3-5 +(axb) (c + 1 - 3). axb3 9 t 12 40x2303 I=8 ( 40 + (40 x230) 16205-230 2 12 2 + qoh Xosz on + 087) (onx 052) + 30+ 40 2 9.291 12

I = 2 x 10 2 x 108 mmy Shear force , V = ? Wok-T z : VQ I where q e Mf Vf S so, nf VF S v[caxb) (c + 1 [(asx uo) (200 + 40 - 1625) I 2x21x103 115 2x108 v = 83,478 N. 1. Maximum shear force on member(a) = 83.478KN

For cross section of member(b) Centroid, y & Ay ΣΑ ý + dx b) le-b) + 2(b xe) + (lxb) (40X270) x 270 +) (170X40)/(270-40) +40 toxud)/(0e70-48) 44 ㄷ 2 (40x270) 170 x 40 cel 162.5 mm Z

2 Moment of inertia, I - Ix + Ad 3 I- b xe 2 +(bxe) (9-9) 12 +(dxb) le-b) + + y I=2 40x3202 + (40*7) (1825-47) 1706402 + (noxv) (210-vw) * 1651 + [ :

I = 2.0 x 108 mm 4 Shear force > V = ? VQ q I nf Vf v [(a+b) (le-b) + b S I 2x21x103 v [(170x44] [1270-40) (6270-40)+40 -162-5 z M5 2 x 108 v= 122762 N .: Maximum shear force on member(b) = 122.762kN

Required spacing,s = ? v xl caxb) (c+ 5 - 9 nf Vf S T. 3 2*21* 103 122.762 X10 [(250x40) (220+40-635) S 2 x108 S = 78.20 mm :. Required spacing spacing 278.20 mm