N.W.F.P. University of Engineering and Technology Peshawar Lecture 07: Miscellaneous topics By: Prof Dr. Akhtar Naeem Khan chairciv@nwfpuet.edu.pk 1 Topics to be Addressed Residual Stresses Factors effecting Residual Stresses Remedial measures against Residual Stresses Effect of Residual stresses on tension members CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 2 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 CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 3 Residual Stresses Residual stresses result principally from nonuniform 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 CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 4 Residual Stresses 1. 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. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 5 Residual Stresses 1. 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. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 6 Residual Stresses 1. Thermal Residual Stresses: W /I shape Distribution of Residual-Stress in W section CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 7 Residual Stresses 1. 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. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 8 Residual Stresses 1. 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 CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 9 Residual Stresses 2. Residual Stresses caused by cold straightening Two straightening procedures: 1. Rotorizing: continuous straightening procedure • Residual stress distribution changes along entire length of member. 2. Gagging: Concentrated straightening at few points • Almost no change in thermal R.S CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 10 Residual Stresses Residual Stresses in 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. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 11 Factors affecting Residual Stresses 1. Geometry 2. Method of preparation 3. Fabricating operation CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 12 Factors affecting Residual Stresses 1. 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 CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 13 Factors affecting Residual Stresses 2. Method of preparation UM Plate CE-409: Lecture 07 Flame-cut Plate Prof. Dr Akhtar Naeem Khan 14 Factors affecting Residual Stresses 2. Method of preparation H from UM Plates CE-409: Lecture 07 H from Flame-cut Plates Prof. Dr Akhtar Naeem Khan 15 Factors affecting Residual Stresses 2. 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. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 16 Factors affecting Residual Stresses 2. Method of preparation Box figures CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 17 Factors affecting Residual Stresses 2. Method of preparation CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 18 Factors affecting Residual Stresses 3. 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. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 19 Remedial Measures against RS 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. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 20 Remedial Measures against RS 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. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 21 Remedial Measures against RS Quenching and Tempering • Because they are quenched and tempered, A514 rolled steel shapes are partially stressrelieved, so residual stresses are small CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 22 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 CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 23 Effect of RS on Tension Members The section is an idealized (web less) H Residual-stress distribution is considered as linear CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 24 Effect of RS on Tension Members CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 25 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 A CE-409: Lecture 07 B Prof. Dr Akhtar Naeem Khan 26 Effect of RS on Tension Members CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 27 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 A CE-409: Lecture 07 B Prof. Dr Akhtar Naeem Khan 28 Effect of RS on Tension Members For fig (h): favg = 36ksi → This gives point C in on stress-strain curve C A CE-409: Lecture 07 B Prof. Dr Akhtar Naeem Khan 29 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 CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 30 Net Section Holes for bolts or rivets in tension members effect the member in two ways 1. Reduce area of x-section 2. Result in non uniform strain on xsection in neighborhood of the hole CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 31 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) CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 32 Net Section: Staggered bolts CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 33 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) CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 34 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. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 35 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. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 36 Net Section Thus the net area for the failure path is given as: s 2t An Ag dt 4g If the plate thickness is uniform, we can divide each term by ‘t’ to get: s2 wn wg d 4g The net width concept is useful when elements of uniform thickness are being evaluated → It is called s2t/4g rule. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 37 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. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 38 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. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 39 Net Section Example 01 Compute the smallest net area for the plate shown below: The holes are for 1 in. diameter bolts. i 3 in. j a b 5 in. c 5 in. 3 in. f h d e 3 in. 3 in. 3 in. 3 in. 3 in. 3 in. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 40 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 CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 41 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. (a) 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 CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 42 Net Section Example 02 CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 43 Net Section Example 02 (a) The fastener gages shown are the usual values for 7x4 angle. s2 wn wg d 4g For section abde: Wn = (7+4-0.75)-2(.75+1/8) = 12 CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 44 Net Section Example 02 (a) The fastener gages shown are the usual values for 7x4 angle. s2 wn wg d 4g For the section abcde: wn = 10.25-(3x0.875)+s2/(4x3) +s2/(4x4.25) CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 45 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 CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 46 Net Section Example 02 (b) The net area is given as: s 2t An Ag dt 4g An = {10.25 - 3x0.875 + 22/(4x3) + 22/(4x4.25)}x0.75 = 6.15in2 CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 47 Net Section Assignment Example 3-6-2 and Example 3-6-3 ( From Gaylord) Due Date: CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 48 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. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 49 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. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 50 Stresses on Net Section According to theory of elasticity, the distribution of stress on net section of infinitely wide plate containing a hole at its centerline is given by f 1 = stress that would exist if there were no hole r = radius of hole x = distance from center of hole to any point on the transverse section. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 51 Stresses on Net Section As the disturbance in stress is highly localized, this equation can be applied with good accuracy to a plate of finite width. It is valid only if the stress at the edge of the hole does not exceed the proportional limit. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 52 Stresses on Net Section Example: Let the plate shown above is subjected to a uniformly distributed tension of 12ksi as shown: It is noted that the stress at the edge of the hole is equal to the yield stress of A36 steel. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 53 Stresses on Net Section Tension members do not always develop a net section average stress equal to tensile strength. Reduction in strength can be expressed in terms of efficiency of net-section. Net-section efficiency is the ratio of average stress at fracture to coupon strength. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 54 Net Section efficiency Net-section efficiency depends upon 1. Ductility of metal 2. Method of making holes 3. Ratio of gage g to fastener diameter d 4. Ratio of net area in tension to area in bearing on fastener(Called bearing ratio) 5. Distribution of x-sectional material relative to gusset plates or other elements to which member is connected. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 55 Net Section efficiency 1.Ductility of metal Because of non uniform stress distribution efficiency of net section is dependent on ductility of metal. Net section in highly ductile material may be 15 to 20 % stronger than same section in material with relative low ductility (from test results) This effect can be expressed by net section efficiency coefficient K1 K1=0.82 + 0.0032R <= 1 CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 56 Net Section efficiency 1.Ductility of metal R: percent reduction in the area of a standard test coupon (2in gage length) R = 50% or more for A36 steel → K1 = 0.98 ≈ 1 → A36 steel is 100% efficient as regards ductility CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 57 Net Section efficiency 2. Method of making holes Punched holes may reduce efficiency of net section by as much as 15% compared with drilled holes. This effect can be expressed by efficiency coefficient K2. K2 = 0.85 for punched holes K2 = CE-409: Lecture 07 1 for drilled, sub-punched and reamed holes Prof. Dr Akhtar Naeem Khan 58 Net Section efficiency 3. g/d ratio Net section is more efficient if ratio of gage g to diameter d is small than if it is large. Efficiency coefficient K3 is proposed for this effect. K3 = 1.6 – 0.7 (An / Ag) CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 59 Net Section efficiency 3. g/d ratio CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 60 Net Section efficiency 4. Net area in tension to area on bearing ratio In case of a plate with uniform gage spacing, the bearing ratio and the fastener spacing are directly related: An ( g d )t g 1 Ag dt d CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 61 Net Section efficiency 4. Net area in tension to area on bearing ratio Tests suggested that strength of connection is not impaired by bearing pressure so long as adequate spacing is provided to produce usual failure modes. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 62 Net Section efficiency 5. Shear Lag effect The phenomenon of non-uniform straining of web. Ends of web are free and four forces shows resultants of bolt shear. Results in shear deformation and stress in web is said to lag because of it. CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 63 Net Section efficiency 5. Shear Lag effect _ k4 x 1 L L: length of the connection i.e. distance from the first fastener to the last one CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 64 Net Section efficiency Effective Net area The effective area is obtained by multiplying the net area with all the above mentioned coefficients: Ae = K1 K2 K3 K4 An CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 65 Thanks CE-409: Lecture 07 Prof. Dr Akhtar Naeem Khan 66