Upcoming SlideShare
×

# Engineering science lesson 5

2,320 views

Published on

Engineering science

Published in: Education
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

### Engineering science lesson 5

1. 1. Chapter 1- Static engineering systems 1.1 Simply supported beams 1.1.1 determination of shear force 1.1.2 bending moment and stress due to bending 1.1.3 radius of curvature in simply supported beams subjected to concentrated and uniformly distributed loads 1.1.4 eccentric loading of columns 1.1.5 stress distribution 1.1.6 middle third rule 1.2 Beams and columns 1.2.1 elastic section modulus for beams 1.2.2 standard section tables for rolled steel beams 1.2.3 selection of standard sections (eg slenderness ratio for compression members, standard section and allowable stress tables for rolled steel columns, selection of standard sections) 1
2. 2. Stresses in beams• Stresses in the beam are functions of x and y• If we were to cut a beam at a point x, we would find a distribution of direct stresses σ(y) and shear stresses σxy(y)• Summing these individual moments over the area of the cross-section is the definition of the moment resultant M,• Summing the shear stresses on the cross-section is the definition of the shear resultant V,• The sum of all direct stresses acting on the cross-section is known as N, 2
3. 3. • Direct stress distribution in the beam due to bending• Note that the bending stress in beam theory is linear through the beam thickness. The maximum bending stress occurs at the point furthest away from the neutral axis, y = c 3
4. 4. Flexure formula•   Stresses calculated from the flexure formula are called bending stresses or flexural stresses.              • The maximum tensile and compressive bending stresses occur at points (c1 and c2) furthest from the neutral surface• where S1 and S2 are called section moduli (units: in3, m3) of the cross- sectional area. Section moduli are commonly listed in design 4 handbooks
5. 5. Euler’s Formula for Pin-Ended Beams v v v v v l Putting l v 5
6. 6. 6
7. 7. 7
8. 8. 8
9. 9. Design of columns under centric loads • Experimental data demonstrate - for large Le/k σcr follows  le /r, (le/k)2 Euler’s formula and depends  upon E but not σY. - for small L/k σcr is  le e/r, determined by the yield  strength σY and not E. - for intermediate Le/k σcr  le /r, depends on both σY and E.   9
10. 10. • For Le/r > Cc l e/k Structural Steel π 2E σ σ cr = σ all = crAmerican Inst. of Steel Construction ( Le/kr ) 2 FS l / FS = 1.92 l e/k • For Le/r > Cc  ( Le /kr ) 2  le / σ σ cr = σ Y 1 − 2  σ all = cr   2Cc   FS 3 5 3 Le/kr 1  Le/k  l / l /r FS = + e −  e  3 8 Cc 8  Cc    le/k • At Le/k = Cc le /r 2 2 2π E σ cr = 1 σ Y Cc = 2 σY 10
11. 11. Sample problem SOLUTION: • With the diameter unknown, the  slenderness ration can not be evaluated.   Must make an assumption on which  slenderness ratio regime to utilize. • Calculate required diameter for  assumed slenderness ratio regime. • Evaluate slenderness ratio and verify  initial assumption.  Repeat if Using the aluminum alloy2014-T6,  necessary.determine the smallest diameter rod which can be used to support the centric load P = 60 kN if  a) L = 750 mm,  b) L = 300 mm 11
12. 12. • For L = 750 mm, assume L/r > 55 • Determine cylinder radius: P 372 × 103 MPa σ all = = A ( L r)2 60 × 103 N 372 × 103 MPa 2 = 2 c = 18.44 mm πc  0.750 m     c/2  • Check slenderness ratio assumption:c = cylinder radius L L 750mmr = radius of  gyration = = = 81.3 > 55 r c / 2 (18.44 mm ) I πc 4 4 c assumption was correct = = 2 = A πc 2 d = 2c = 36.9 mm 12
13. 13. • For L = 300 mm, assume L/r < 55• Determine cylinder radius: P   L  σ all = = 212 − 1.585  MPa A   r  60 × 103 N   0.3 m  6 = 212 − 1.585  × 10 Pa πc 2   c / 2  c = 12.00 mm• Check slenderness ratio assumption: L L 300 mm = = = 50 < 55 r c / 2 (12.00 mm ) assumption was correct d = 2c = 24.0 mm 13
14. 14. Eccentric loading of columns• Generally, columns are designed so that the axial load is inline with the column• There are situations that the load will be off center and cause a bending in the column in addition to the Pin-Pin Column  compression. This type of loading is called eccentric load with Eccentric  Axial Load • When a column is load off center, bending can be sever problem and may be more important than the compression stress or buckling 14
15. 15. Analysis of eccentric loads• At the cut surface, there will be both an internal moment, m, and the axial load P. This partial section of the column must still be equilibrium, and moments can be summed at the cut surface, giving, ΣM = 0 m + P (e + v) = 0• bending in a structure can be modeled as m = EI d2v/dx2, giving EI d2v/dx2 + Pv = -Pe• This is a classical differential equation that can be solved using the general solution, v = C2 sin kx + C1 cos kx - e where k = (P/EI)0.5. The constants C1 and C2 can be determined using the boundary conditions 15
16. 16. • First, the deflection, v=0, at x = 0 0 = C2 0 + C1 1 - e C1 = e• The second boundary condition specifies the deflection, v=0, at X = L 0 = C2 sin kL + e cos kL - e C2=e tan (kL/2)• Maximum deflection – The maximum deflection occurs at the column center, x = L/2, since both ends are pinned. 16
17. 17. Maximum stress: secant formula• Unlike basic column buckling, eccentric loaded columns bend and must withstand both bending stresses and axial compression stresses.• The axial load P, will produce a compression stress P/A. Since the load P is not at the center, it will cause a bending stress My/I.• The maximum moment, Mmax, is at the mid-point of the column (x = L/2), Mmax = P (e + vmax) 17
18. 18. • Combining the above equations gives• But I = Ar2. This gives the final form of the secant formula as• The stress maximum, σmax, is generally the yield stress or allowable stress of the column material, which is known.• The geometry of the column, length L, area A, radius of gyration r, and maximum distance from the neutral axis c are also known. The eccentricity, e, and material stiffness, E, are considered known. 18
19. 19. 19
20. 20. Design of columns under an eccentric load • An eccentric load P can be replaced by a  centric load P and a couple M = Pe. • Normal stresses can be found from  superposing the stresses due to the  centric load and couple, σ = σ centric + σ bending P Mc σ max = + A I • Allowable stress method: P Mc + ≤ σ all A I • Interaction method: P A Mc I + ≤1 ( σ all ) centric ( σ all ) bending 20
21. 21. Example The uniform column consists of an 8-ft section  of structural tubing having the cross-section  shown. a) Using Euler’s formula and a factor of safety  of two, determine the allowable centric load  for the column and the corresponding  normal stress. b) Assuming that the allowable load, found in  part a, is applied at a point 0.75 in. from the  geometric axis of the column, determine the  horizontal deflection of the top of the  column and the maximum normal stress in  the column. 21
22. 22. SOLUTION:• Maximum allowable centric load:- Effective length, Le = 2( 8 ft ) = 16 ft = 192 in.- Critical load, Pcr = π 2 EI = ( )( π 2 29 × 106  psi 8.0 in 4 ) 2 Le (192 in ) 2 = 62.1 kips- Allowable load, P 62.1 kips Pall = 31.1 kips Pall = cr = FS 2 P 31.1 kips σ = all = σ = 8.79 ksi A 3.54 in 2 22
23. 23. • Eccentric load: - End deflection,  π P   ym = e sec   2 P  − 1   cr     π   = ( 0.075 in ) sec  − 1  2 2  ym = 0.939 in. - Maximum normal stress, P  ec  π P  σm = 1 + 2 sec  2 P   A r  cr   31.1 kips  ( 0.75 in )( 2 in )  π  = 2  1+ sec  3.54 in  (1.50 in ) 2  2 2  σ m = 22.0 ksi 23
24. 24. ExampleDetermine the maximum flexural stress produced by a resisting Moment Mr of+5000ft.lb if the beam has cross section shown in the figure. Locate the neutral axis from the bottom end 24
25. 25. 25
26. 26. • Work out the rest of example here 26