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By: Prof Dr. Akhtar Naeem Khan chairciv@nwfpuet.pk

Lecture 07: Miscellaneous topics . By: Prof Dr. Akhtar Naeem Khan chairciv@nwfpuet.edu.pk. Topics to be Addressed. Residual Stresses Factors effecting Residual Stresses Remedial measures against Residual Stresses Effect of Residual stresses on tension members. Residual Stresses.

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By: Prof Dr. Akhtar Naeem Khan chairciv@nwfpuet.pk

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  1. Lecture 07: Miscellaneous topics By: Prof Dr. Akhtar Naeem Khan chairciv@nwfpuet.edu.pk

  2. Topics to be Addressed • Residual Stresses • Factors effecting Residual Stresses • Remedial measures against Residual Stresses • Effect of Residual stresses on tension members

  3. Residual Stresses • Tension members response to load is much • similar to that of tensile-test coupon . • Member behavior may differ from coupon • behavior because of: • Slip in bolted & riveted connections • Non-linear behavior of connections • Residual stresses in member

  4. Residual Stresses • Residual stresses result principally from non-uniform cooling of hot rolled or welded shapes and from cold straightening of bent members. • 1. Thermal Residual Stresses • 2. Residual Stresses caused by cold straightening

  5. Residual Stresses Thermal Residual Stresses: W /I shape • As they have more surface exposure per unit of volume, flange tips and central parts of the webs tends to cool faster than juncture of flange-web of a section (w-f) i.e. rate of cooling of juncture is slower than rate of cooling of tips • As a result, metal at junctures continues to contract as it cools after flange tips and web interior have cooled to temperature of surroundings.

  6. Residual Stresses Thermal Residual Stresses: W /I shape • This contraction is partially restrained by cooler metal which causes: • Tensile stresses to develop at the juncture of flange-web • Compressive stresses in the remainder of the cross-section • These are called residual Stresses.

  7. Residual Stresses Thermal Residual Stresses: W /I shape Distribution of Residual-Stress in W section

  8. Residual Stresses Thermal Residual Stresses: W /I shape 20 W’s shapes were investigated: • It Revealed that flange-tip stress frc varied from 4.1 to 18.7 Ksi, the average being 12.8 ksi • Residual stresses in web center varied from 41Ksi compression to 18.2Ksi tension. • Showing some W’s develop residual tension over entire web, instead the pattern shown.

  9. Residual Stresses Thermal Residual Stresses: W /I shape • Only one out of 20 sections was thicker than 1 in. • Therefore, above values are not representative of W’s with thick flanges and webs • Residual stresses tend to increase in magnitude with increase in thickness

  10. Residual Stresses Residual Stressescaused by cold straightening • Two straightening procedures: • 1. Rotorizing: continuous straightening procedure • Residual stress distribution changes along entire length of member. • Gagging: Concentrated straightening at few points • Almost no change in thermal R.S

  11. Residual Stresses Residual Stressesin welded connection • Because of high concentration of heat, tensile residual stresses at the weld in welded members usually equal the yield strength of the weld metal itself which may be as much as 50% higher than that of the parent metal.

  12. Factors affecting Residual Stresses Geometry Method of preparation Fabricating operation

  13. Factors affecting Residual Stresses Geometry • Magnitude and distribution of thermal residual stresses are influenced to considerable degree by geometry of x-section • Residual stresses tend to increase in magnitude with increase in thickness

  14. Factors affecting Residual Stresses Method of preparation UM Plate Flame-cut Plate

  15. Factors affecting Residual Stresses Method of preparation H from UM Plates H from Flame-cut Plates

  16. Factors affecting Residual Stresses Method of preparation • Large residual stresses develop at the corners of the welded box • On the other hand, residual stresses in the hot-rolled square box are very low and in one investigation averaged less than 5 ksi.

  17. Factors affecting Residual Stresses Method of preparation • Box figures

  18. Factors affecting Residual Stresses Method of preparation

  19. Factors affecting Residual Stresses Fabricating operation • Fabricating operations such as cambering and straightening by cold bending also induce residual stresses. • These are of about the same magnitude but differ in distribution • These stresses are superimposed on the thermal residual stresses.

  20. Remedial Measures against R S Quenching and Tempering • Quenching is the act of rapidly cooling the hot steel to harden the steel. • Quenched steel is hard and brittle. • Often it is just too brittle and must be made more malleable, This is achieved by a process known as tempering.

  21. Remedial Measures against R S Quenching and Tempering • The quenched steel is heated again but this time to a temperature between 200 °C and 300 °C. • When the metal reaches the tempering temperature, it is quenched again in cold water or oil. The result is a steel that is still hard but is more malleable and ductile.

  22. Remedial Measures against R S Quenching and Tempering • Because they are quenched and tempered, A514 rolled steel shapes are partially stress-relieved, so residual stresses are small

  23. Effect of RS on Tension Members • Generally tension members response to load is much similar to that of tensile-test coupon but not identical. • However member behavior may differ from coupon behavior because of: • Slip in bolted & riveted connections • Non-linear behavior of connections • Residual stresses in member

  24. Effect of RS on Tension Members • The section is an idealized (web less) H • Residual-stress distribution is considered as linear

  25. Effect of RS on Tension Members

  26. Effect of RS on Tension Members • For Fig (d): • P = 2(24x12x1) = 576kips • favg = 576/24 = 24ksi • → This gives point A on the stress-strain curve C B A

  27. Effect of RS on Tension Members

  28. Effect of RS on Tension Members • For fig (f): • P = 2(36x6x1 + 30x6x1) = 792kips • favg = 792/24 = 33ksi • → This gives point B on stress-strain curve C B A

  29. Effect of RS on Tension Members • For fig (h): • favg = 36ksi • → This gives point C in on stress-strain curve C B A

  30. Effect of RS on Tension Members • No effect on the yield strength of the member • Lowering of proportional limit (P.L <36ksi) • Increase in the strain at initiation of overall yielding • →No consequence in regard to the static strength of the member • →Can be important if fatigue is involved • →R.S have a pronounced effect on the strength of columns

  31. Net Section • Holes for bolts or rivets in tension members effect the member in two ways • Reduce area of x-section • Result in non uniform strain on x-section in neighborhood of the hole

  32. Net Section • Net area is defined as gross section minus area which is lost because of holes. • Effective net area is obtained by multiplying net area by coefficient to account for its reduced effectiveness if not all the member elements are connected. (According to AISC)

  33. Net Section: Staggered bolts

  34. Net Section: Staggered bolts • Failure paths may occur on sections normal to axis of member (1-2-5) or may include zigzag sections (1-2-3-4). • Depending on the relative values of g, s and bolt diameter d. g: gauge (distance btw longitudinal fastener line) s: pitch (distance btw transverse rows)

  35. Net Section: Staggered bolts • If g > s : failure is expected along 1-2-3-4 • If g < s : failure is expected along 1-2-5 • For fixed values of g and s…. failure is expected along the zigzag section as the size of holes increases.

  36. Net Section • Empirical methods have been developed to calculate the net section fracture strength • Assumption: The effect of the zigzags in any failure path can be accounted for by deducting from the area of the section the areas lost by the holes in the failure path and adding the quantity (s2t/4g) for each zigzag.

  37. Net Section Thus the net area for the failure path is given as: If the plate thickness is uniform, we can divide each term by ‘t’ to get: The net width concept is useful when elements of uniform thickness are being evaluated → It is called s2t/4g rule.

  38. Net Section Staggered bolts in angles. • If staggered lines of bolts are present in both legs of an angle, then the net area is found by first unfolding the angle to obtain an equivalent plate. • The unfolding is done at the middle surface to obtain a plate with gross width equal to the sum of the leg lengths minus the angle thickness.

  39. Net Section Staggered bolts in angles. • AISC Specification B2 says that any gage line crossing the heel of the angle should be reduced by an amount equal to the angle thickness. • For this situation, the distance g will be = 3 + 2 – ½ in.

  40. Net Section Example 01 Compute the smallest net area for the plate shown below: The holes are for 1 in. diameter bolts.

  41. Net Section Example 01 The effective hole diameter is 1 + 1/8 = 1.125 in. For line a-b-d-e wn = 16.0 – 2 (1.125) = 13.75 in. For line a-b-c-d-e wn = 16.0 – 3 (1.125) + 2 x 32/ (4 x 5) = 13.52 in. The line a-b-c-d-e governs: An = t wn = 0.75 (13.52) = 10.14 in2

  42. Net Section Example 02 • Example 3-6-1. A 7x4x3/4 angle is connected by two rows • of 3/4in Bolts in the 7in leg and one row in the 4in leg • as shown. Standard holes are used. • Determine the pitch s so that only two holes for 3/4in • Fasteners need to be deducted in computing the net area • (b) Determine the net area of the 7x4 angle if the pitch s • is 2in

  43. Net Section Example 02

  44. Net Section Example 02 (a) The fastener gages shown are the usual values for 7x4 angle. For section abde: Wn = (7+4-0.75)-2(.75+1/8) = 12

  45. Net Section Example 02 (a) The fastener gages shown are the usual values for 7x4 angle. For the section abcde: wn = 10.25-(3x0.875)+s2/(4x3) +s2/(4x4.25)

  46. Net Section Example 02 (a) For section abde: Wn = (7+4-0.75)-2(.75+1/8) = 12 For the section abcde: wn = 10.25-(3x0.875)+s2/(4x3) +s2/(4x4.25) Equating and solving for s we get: S = 2.48in ≈ 2.50in An = (10.25-3x0.875)0.75 = 6.38in2

  47. Net Section Example 02 (b) • The net area is given as: An = {10.25 - 3x0.875 + 22/(4x3) + 22/(4x4.25)}x0.75 = 6.15in2

  48. Net Section Assignment • Example 3-6-2 and Example 3-6-3 ( From Gaylord) Due Date:

  49. Stresses on Net Section Consider the non uniform strain in the vicinity of the hole in a uniformly Stretched sheet of rubber as shown → The unloaded sheet upon which an orthogonal grid is drawn.

  50. Stresses on Net Section The stretched sheet The strains at the edge of the elongated hole are much larger than those elsewhere in the sheet. The disturbance is highly localized.

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