Presentation is loading. Please wait.

Presentation is loading. Please wait.

INTRODUCTION Department Of Civil Engineering Govt. Poly. College, Bathinda Topic – Tension member Subject – Steel Structures Design Presenter- Er. Rani.

Similar presentations


Presentation on theme: "INTRODUCTION Department Of Civil Engineering Govt. Poly. College, Bathinda Topic – Tension member Subject – Steel Structures Design Presenter- Er. Rani."— Presentation transcript:

1 INTRODUCTION Department Of Civil Engineering Govt. Poly. College, Bathinda Topic – Tension member Subject – Steel Structures Design Presenter- Er. Rani Devi B.E.(Civil), M.E. (Structures) Mob.- 9465265746 12th March, 2013Punjab Edusat Society1

2 Introduction Steel structures are structures in which the members are made of steel and are joined by welding, riveting, or bolting. Because of the high strength of steel, these structures are reliable and require less material than other types of structures. In modern construction, steel structures are widely used such as industrial buildings, storage tankgas tanks, communication structures (radio and television towers and antennas), and power-engineering structures. 12th March, 2013Punjab Edusat Society2

3 The top beams in a truss are called top chords and are generally in compression, the bottom beams are called bottom chords and are generally in tension, the interior beams are called webs, and the areas inside the webs are called panels. 12th March, 2013Punjab Edusat Society 3

4 12th March, 2013Punjab Edusat Society4 Design for the steel frame of a twin-bay industrial building: (1) lattice, (2) column, (3) crane girder, (4) skylight, and (5) web members

5 Tension Members Members which are subjected to direct tension are called tension members. The Members can be of any standard steel section e.g., angle iron, channel section etc. In case of frames and trusses, a tension member is called a tie. 12th March, 2013Punjab Edusat Society5

6 Types of Sections The types of section is governed by the nature and magnitude of stresses to which, it is subjected. The different type of sections are shown in figure and classified as: (1)Rod, round or square :- These are used in buildings for the lateral and sway bracings, hangers, segmental arch floors and timber trusses etc. 12th March, 2013Punjab Edusat Society6

7 (2) Flats :- These are used as tension member in light trusses connected by welding at their ends. (3) Eye bars :- These are used in pin connected Structures. 12th March, 2013Punjab Edusat Society7

8 (4) Single angle sections :- These are used as tension member in light roof trusses, bracing members in plate girder bridges and light latticed girder bridges, not recommended as best tension member as they are subjected to bending stresses due to the eccentric loads. 12th March, 2013Punjab Edusat Society8

9 (5) Double angle section :- These are used extensively in roof trusses, may be connected to gusset plate on same or opposite faces i.e., by placing the gusset plate in between the two angles. The later type of end connection is preferred and more economical as it eliminates bending stresses. 12th March, 2013Punjab Edusat Society9

10 12th March, 2013Punjab Edusat Society10

11 (6) T-sections :- These are used as a substitute of former type of double angles. (7) Double channel section :- These are used for heavy structures subjected to bending and direct stresses. 12th March, 2013Punjab Edusat Society11

12 Strength of a Tension Member The Strength of a Tie Member or load bearing capacity of a tension member is calculated as :- Strength of a Tension Member = Net area × Permissible tensile stress(σ at ). The strength depends upon net area. Permissible tensile stress(σ at ) for a tension member is generally taken as 150MPa 12th March, 2013Punjab Edusat Society12

13 Net Sectional Area Net sectional area of a tension member is the gross cross section area of the member minus the deduction for holes. Case 1 - For Plate : In determining the net sectional area of the plates, the arrangement of the rivets plays an important role. The riveting in plate can be of two types i.e.: (a)Chain riveting (b) Zig-Zag riveting. 12th March, 2013Punjab Edusat Society13

14 (a) Plates connected by Chain riveting : Let, A net = Net cross sectional area along any section 1-1 b = Width of Plate n = number of rivets along the section under consideration 12th March, 2013Punjab Edusat Society14

15 d = Gross diameter of the rivets t = Thickness of the plate A net = Gross area- area of plate lost in the process of making holes A net = b t - nd t A net = (b – nd) t 12th March, 2013Punjab Edusat Society15

16 (b) Plates connected by Zig-Zag or staggered riveting : Let, A net = Net area of cross sectional b = Width of Plate n = number of rivets in the plane under consideration. 12th March, 2013Punjab Edusat Society16

17 n = number of gauge distance s = staggered pitch g = gauge distance t = Thickness of the plate A net = 12th March, 2013Punjab Edusat Society17

18 Case 2 - For single angle connected by one leg only: Net effective area, i.e., A net = A 1 + A 2 K Where, A net = Net cross-sectional area A 1 = Net cross-sectional area of connected leg 12th March, 2013Punjab Edusat Society18

19 A 1 = (length of leg – ½ t – nd) × t A 2 = Gross cross-sectional area of outstanding leg A 2 = (length of leg – ½ t) x t K = constant = A net = A 1 + A 2 K 12th March, 2013Punjab Edusat Society19

20 Case 3. For a pair of angles place over a single tee connected by only one leg of each angle to the same side of the gusset plate : 12th March, 2013Punjab Edusat Society20

21 Net effective area, i.e., A net = A 1 + A 2 K Where, A net = Net cross-sectional area A 1 = Net cross-sectional area of connected leg (flange of the tee) A 2 = Gross cross-sectional area of outstanding leg (web of the tee) K = constant = 12th March, 2013Punjab Edusat Society21

22 12th March, 2013Punjab Edusat Society22 A 1 = 2 (length of leg – ½t – nd) × t A 2 = 2 (length of leg – ½ t) × t In case of T section A 1 = (b f - 2d ) × t f A 2 = Gross area of web = t w × (overall depth(h) – t f )

23 Case 4. For double angles or a tee placed back to back and connected to each side of the gusset or side of a rolled section. :- 12th March, 2013Punjab Edusat Society23

24 Deduction for rivet holes = number of rivets × gross diameter of one rivet × thickness of plate i.e., = n × d × t Net effective area, A net = Gross area – Deduction for rivet holes 12th March, 2013Punjab Edusat Society24

25 Problem 1: A plate 240mm wide and 12mm thick is connected by 20mm φ rivets as shown. Calculate the strength of the tension plate. Take σ at = 150N/mm 2 12th March, 2013Punjab Edusat Society25 Numerical Problems

26 Solution: Nominal dia. of rivet (D) =20mm Gross dia. of rivet hole (d) = 20 +1.5 = 21.5mm Width of plat = 30 + 60 + 60+ 60 + 30 = 240mm s = 50mm g = 60mm Considering the failure of the plate along section line 1-2-3-4, we have; A net 1 = (b – nd) t t = 12mm, n = 2, b = 240mm, d = 21.5mm 12th March, 2013Punjab Edusat Society26

27 A net 1 = (240 -2 × 21.5) × 12 = 3564mm 2 Now for area along staggered line 1-2-5-3-6-7 A net 2 = Where b = 240mm n = 4 d = 21.5mm n = 3 t = 12mm 12th March, 2013Punjab Edusat Society27

28 A net 2 = = 2073 mm 2 Minimum net area of section = A net 2 Strength of the plate = 2073 × 150 = 310950 = 310.95kN 12th March, 2013Punjab Edusat Society28

29 Problem 2: Calculate the strength of a tie composed of ISA 100 × 75 × 8mm with longer legs connected by 16mm dia. rivets. Take σ at = 150N/mm 2 12th March, 2013Punjab Edusat Society29

30 Solution: Nominal dia. of rivet = 16mm Gross dia. of rivet = 16 + 1.5 = 17.5mm Net Area; A net = A 1 + A 2 K A 1 = = 628mm 2 A 2 = 561mm 2 K= constant = 12th March, 2013Punjab Edusat Society30

31 A net = A 1 + A 2 K = 628 + 561 × 0.77 = 1059.97mm 2 Strength of ISA 100 × 75 × 8 = 1059.97mm 2 × 150 = 158995.5N = 158.99kN 12th March, 2013Punjab Edusat Society31

32 Problem 3: Calculate the strength of a tie member composed of 2 ISA 125 × 75 × 8mm placed back to back connected by longer legs by 20mm dia. rivets on same side of the gusset plate. Take σ at = 150N/mm 2 12th March, 2013Punjab Edusat Society32

33 Solution: Nominal dia. of rivet (D) =20mm Gross dia. of rivet hole (d) = 20 +1.5 = 21.5mm A net = A 1 + A 2 K A 1 = = 1592mm 2 A 2 = = 1136mm 2 K = constant = 12th March, 2013Punjab Edusat Society33

34 K = Net effective area A net = A 1 + A 2 K = 1592 + 1136 × 0.875 = 2586mm 2 Strength of tie member = 2586 × 150 = 387900N = 387.9kN 17th April,2012Punjab Edusat Society34 ×

35 Problem 4: Calculate the strength of a tie member composed of 2ISA 125 × 75 × 8mm placed back to back connected by longer legs by 20mm dia rivets on both sides of the gusset plate. 12th March, 2013Punjab Edusat Society35

36 Solution: From steel table, gross area of ISA 125 × 75 × 8mm = 15.38cm 2 = 1538mm 2 Net effective area = Gross area – Deduction for rivet holes A net = 2[1538 – 21.5 × 8] = 2732mm 2 Strength of tie member = 2732 × 150 = 109.8kN 12th March, 2013Punjab Edusat Society36

37 Problem 5: Determine the strength of ISHT 75 which is used as a tie member. It is connected through its flange by means of 20mm diameter rivets. Take σ at = 150N/mm 2 Solution: Data Given: Tie member consists of ISHT 75 Nominal dia. of rivet (D) =20mm Gross dia. of rivet hole (d) = 20 +1.5 = 21.5mm 12th March, 2013Punjab Edusat Society37

38 Thickness of Flange, t w = 8.4mm Thickness of web, t f = 9mm Overall Height = 75mm 12th March, 2013Punjab Edusat Society38

39 Case under consideration : Tee (or 2ISA) connected on same side of the gusset plate by riveting: A net = A 1 + A 2 K K = constant = Properties of ISHT 75 (From steel cables) Width of plate, b = 150mm 12th March, 2013Punjab Edusat Society39

40 Net Area of flange, A 1 = (width of flange –nd) × t f = (150 – 2 × 21.5) × 9 (Since, number of rivets(n) = 2) = 963mm 2 Gross area of web, A 2 = (75 - 9) × 8.4 (Since, Area = Length of web × t w ) = 554.4mm 2 K = 12th March, 2013Punjab Edusat Society40

41 Net cross sectional area, A net = A 1 + A 2 K = 963 + 554.4 × 0.897 = 1460.29mm 2 Tensile Strength of the T – section = A net × σ at = 1460.29 × 150 = 219043.5N = 219.04kN 12th March, 2013Punjab Edusat Society41

42 Design of Members Subjected to Axial Tension Step 1. Calculation of the required net area The axial pull (force) to be transmitted by the member and the allowable stress in axial tension ( permissible tensile force i.e. σ at ) are known for the steel with yield stress f y. Net sectional area required = 12th March, 2013Punjab Edusat Society42

43 i.e. A net reqd. = Step 2. Selection of suitable section Try a suitable section, from steel tables having sectional area about 20 to 40% greater (in case of riveted joint) and 10% greater (in case of welded joint) than the required net area. In case the member selected is ISA, than select unequal angle and connect longer leg with gusset plate for getting more strength i.e. (load carrying capacity) 12th March, 2013Punjab Edusat Society43

44 Step 3. Calculate the net sectional (effective) area of the selected section (As discussed earlier) Step 4. Check for net Sectional area The net area calculated for trial section (in step 3) should be slightly greater than the required net area. If it is so then selected section is OK. Other wise try some other section. i.e. Net area of trial section > Net area required 12th March, 2013Punjab Edusat Society44

45 Step 5. Check for Slenderness ratio (λ). Slenderness ratio, is the ratio of effective length of member to the least radius of gyration. i.e. Slenderness ratio λ = The value of radius of gyration (r min ) can be obtained from steel tables. 12th March, 2013Punjab Edusat Society45

46 1.A tension member in which a reversal of direct stress due to load other than wind or seismic forces would occur, shall not have a slenderness ratio more than 180. 2.A member normally acting as a tie in a roof truss, but subject to possible reversal of stress resulting from the action of wind or seismic forces shall have slenderness ratio not greater than 350. 12th March, 2013Punjab Edusat Society46

47 Step 6. Design of end connections The end connections may be designed as a riveted connection or welded connection. (a)For riveted connection (i)Select suitable size of rivet and determine the rivet value ( i.e. least of P b and P f ) (ii)Find the number of rivets by the relation ; Number of rivets required = 12th March, 2013Punjab Edusat Society47

48 (iii) The arrangement of rivets should be made in such a way that : there is no eccentricity of loading. The centre of gravity of the section coincides with the C.G. of group of rivets. 12th March, 2013Punjab Edusat Society48

49 (b) For welded connections (i)Find the minimum and maximum size of fillet weld and select the suitable size of weld (S) and find the value of effective throat thickness, t = 0.7 S (ii)Calculate the strength of weld/mm length by the formula; = τ vf × l × t 12th March, 2013Punjab Edusat Society49

50 Problems Based On Design of Tension Member Problem 1.Design a tension member subjected to pull of 165 kN using unequal angles placed back to back with their longer legs connected on both sides of gusset plate by 18mm diameter rivets. Use PDSR (Power Driven Shop Rivets). Take σ at = 150N/mm 2 12th March, 2013Punjab Edusat Society50

51 Solution : Data given ; Load (axial pull) = 165kN = 165 × 10 3 N Nominal dia. of rivet, D = 18mm Gross dia. of rivet, d = 18 + 1.5 = 19.5mm For PDSR, τ vf = 100 N/mm 2 σ pf = 300 N/mm 2 σ at = 150 N/mm 2 12th March, 2013Punjab Edusat Society51

52 Net area required = = = 1100mm 2 Assuming the gross area to be about 25% greater than the net area Gross area = 1100 × 1.25 = 1375mm 2 from steel cable, try 2 ISA gross area equal to or greater than 1375mm 2 Let us try an ISA 80 × 50 × 6 mm @ 57.9 N/m = 7.46cm 2 or 746mm 2 12th March, 2013Punjab Edusat Society52

53 Gross area for 2 ISA = 2 × 746 = 1492mm 2 Longer legs are connected to the gusset plate (as shown in Figure) Case under construction : Two ISA connected back to back on the both sides of the gusset plate ( i.e. CASE IV) A net = Gross area - Deduction for holes 12th March, 2013Punjab Edusat Society53

54 2 ISA connected back to back (with longer leg connected) on the both sides of gusset [late A net = 1492 – (2 × 19.5) × 6 = 1258 > 1100mm 2 Hence, use 2 ISA 80 × 50 × 6 mm @ 57.9 N/m as a tension member. 12th March, 2013Punjab Edusat Society54

55 Design of riveted end connections Shearing strength of one rivet (in double shear) = 2 × τ vf × = 2 × 100 × = 59729.53 N...(1) Bearing Strength of one rivet = σ pf × d × t = 300 × 19.5 × 6 = 35100 N …(2) 12th March, 2013Punjab Edusat Society55

56 Least of (1) and (2) is the Rivet Value (R.V.) R.V. = 35100 N Number of rivets required = Arrangements of rivets is as shown in Figure. Provide pitch = 3 × 18 = 54 = 60mm c/c Edge distance = 2 × 18 = 36 = 40mm 12th March, 2013Punjab Edusat Society56

57 12th March, 2013Punjab Edusat Society57

58 CONCLUSION Tension member. Various sections which are used as tension members. Strength of tension members. Different formulas to calculate net effective area for various sections. Design of Tension Members. Problems. Thanks 12th March, 2013Punjab Edusat Society58


Download ppt "INTRODUCTION Department Of Civil Engineering Govt. Poly. College, Bathinda Topic – Tension member Subject – Steel Structures Design Presenter- Er. Rani."

Similar presentations


Ads by Google