Question

3-34 For each section illustrated, find the second moment of area, the location of the neutral axis, and the distances from the neutral axis to the top and bottom surfaces. Consider that the section is transmitting a positive bending moment about the z axis, M., where M. = 10 kipin if the dimen- sions of the section are given in ips units, or M. = 1.13 kNm if the dimensions are in SI units. Determine the resulting stresses at the top and bottom surfaces and at every abrupt change in the cross section. 40 mm 1 25 mm B 25 mm -1 in (b) Problem 3-34 75 100 1 12.5 125 B А 125 25 50 100 (c) Dimensions in mm (d)

Answer 1

Solution Ans(a): – Draw the diagram of the section representing the two sections. 40mm D 3.7.5mm 6 Том К. 75 mm А. N-* 2 12.5mm 25 roron N -> total area of the cross section : A = As A2 75840 25 X34 = 2150 mm² A= 3000 - 850 y To tal moment of Inertia of section:- I= IL - I2 = (40) (75) (60-6) (251 12 12 I= 1406250 - 44270.83 1361979.16 mm 4 = second moment of area I= from the top & bottom s distance of neutral arus is 37.5mm stress at point (A) SA Mz C I A = 1.13x106(H.org X 37. 5(mm) 136 1179.16 (mmy) S6A = 31. 11. N/m 31. 11 MPa 3 (tersise

due to similarity stress at polet :- 16 - 31..11. N Proma? Compressivel i Stress at point :- I. 1 3x 106 (1. mm) X (37.5-25) mm) B Mec MZ C son 10.37 MPaļ (Tensive) Similarly, stress at point C) :- Şoc = 10.37 MPa Mpal (compressive) 1 36 19 79. 16 Ang(6) ly D 1.0itin 0.9875 in GE N-* A V en С. 012- 0,858 in 0.6875 in A V — к 3/8in 13% in K 3/8 in area a Total of the cross section: A=2 Alt Az 2(1.875) (0.375) (1.75) (0.375) A=2.0625 in2 centsold of section , 1+ 3/8+ 1 / 2 = 0.9375 in 9 = 2 centroid of section (@, ge= 1+ 3/8 1 = 0.6875 in Find the controid of the section?

y= I ALI1+ Asya 2(1.875)(0.376%0.9375+(1.75)(0.375)64 mg) 2 A3+ Ae 21.54570.375)+(35) 6-757 J = 0.858 In (from the bottom) &y= 1.875-0.858 (from the top 1.017 in . Moment of inenda of section 1: Id = (0,375) (1.875)S 0,206 inh . Moment of Inertia of section ! 12= (0.3459 (1.75) 0.00769 in 4 - 12 • The total moment of inostia of the section I = 2 (It + Azdł !+ [I2+ A2d?) = 2 [0.206+ 0.371) 68.845) 4.01730.93+5)] + (0.00769 + (1.75)(0.375)(0,858- 0.6875)?] {I 0.44765 Int} = second moment of menta Location of neutral orů u is 1.017sh from the top & 0.858 in from the bottom, . Find the stress at point A M2 C 10 (kips-in X 0.858 0.4465 a = 19.167 kmiy (Tensite) The stress at point B, M₂ C 10 X (0.858-0.5) І 0.44 765 - 7.997 Kesily (Tensile .

The stress 6 с 2 I at point C:- Mac 10 x 60 858+ 0.875) 0.44765 {$C = 0.3797 kpsit (compressives stress at point Di- D Mac 10 X (1.017) 0.44765 22.7186 kpsi] c compressive) . I Anally:- 42.71mm 100 mm 57.29m/ room 125 1125 & 12 as T → Area of the cross section: A = A3-A2-As = (100X100) – (75X75)–(50 X 12.57 10000 - 5625- 625 A = 3750 mm² y = 12.5 o centroid of section Centrold of section ② 12.5 6.25mm y2= 12.5+ 75 somm 2 centroid of section 9 = 100 = somm - 2 Centroid of the section AS Y3 - A242- A191 A T = (1000g)(50) – (5625)(50) – (625)(623) - 57.2 mm 3750

Location of neutral wis is 57.20 from the bottom 0 42.71 from the top. - Moment of inertia section (1): – I1 = (50) (12.53 8138 mm 12 I2 = (75) (788 2.637 X106 mm 12 I3 (100)20013 8.333 8106 mmy 12 • Total I = I = [I 3 + Az d? ] - [I2 + A2d2] - [IL + As dz] [8.333X1oºt (1000X57.29-8013]-[2.637X1067 (1629)657:23-Soy}] -[8138+ (625)(57.29–6.2012] I=4292434 46 mm" = 15:08 MPa (T) - M₂ C 11.79 MPa (T) M2 C (1.13X100) (57.29) 1 4 29 2434.46 (1.13X106) (57.29-12.5) 4292434.46 (1.13 X106)(42. 71-12,5) 4292434.46 01.13X106) (42.71) = 929243 4. 46 = 7o D5 MPa (C) ore = M2c I 13.24 mma (CD obc = Mac I uin Aud): - 0.875 1.087 in 2. Sin 2.288 in A → K..370

- Total area of the section - A = A + A2 = (0.875) (4) + (215) (0,875) = 5.6875 in 3.5 + 2.1875 centroid of 2- the section (1) = 2.5+ (0.875)=2.9375 in (2), 42 = 2of 1.25 in 2 Centroid of seation = ALM+ A2Y А (3.5) (2.9375) + (2.3875) (1.25) 5.6875 ý = 2.288 in from the bottom) a j = 1.087 in (from the top 410.87583 = 0.2233 in 4 12 Is = 9.875) (253 S 1. L 39 in4 12 I = [Ist Agdf] + (Iz +1 2 2 2 ] - 10.2233 +0.5)(1.087-0.4375)[1,10+(2.1875)(2.88-4 88-1.253] I- 5.195 iny M2C 20 X 2.288 =4.404 Kgi (T SA 5.195 M₂C I 10 X (1.087) 2.092 kpsi (C) 5.195 LOK (2.5 2.288) 0.408 Kapsi S. 195 = . 18 M2 C I

Thank you.