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

Punjab Edusat Society 1

INTRODUCTIONDepartment Of Civil EngineeringGovt. Poly. College, Bathinda

Topic – Tension memberSubject – Steel Structures Design

Presenter- Er. Rani DeviB.E.(Civil), M.E. (Structures)

Mob.- 946526574612th March, 2013


IntroductionSteel 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.

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Punjab Edusat Society

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.

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Punjab Edusat Society 412th March, 2013

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


Tension MembersMembers 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.

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Types of SectionsThe 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.

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(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.

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(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.

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(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.

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(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.

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Strength of a Tension MemberThe 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(σat).

The strength depends upon net area.Permissible tensile stress(σat) for a tension member is generally taken as 150MPa

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Net Sectional AreaNet 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.

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(a) Plates connected by Chain riveting :

Let, Anet = Net cross sectional area along any

section 1-1 b = Width of Plate n = number of rivets along the section

under consideration

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d = Gross diameter of the rivets t = Thickness of the plate Anet = Gross area- area of plate lost in the

process of making holes Anet = b t - nd t

Anet = (b – nd) t

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(b) Plates connected by Zig-Zag or staggered riveting :

Let, Anet = Net area of cross sectional

b = Width of Plate n = number of rivets in the plane under

consideration. 12th March, 2013


n’ = number of gauge distance s = staggered pitch g = gauge distance t = Thickness of the plate Anet =

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Case 2 - For single angle connected by one leg only:

Net effective area, i.e., Anet = A1 + A2K

Where, Anet = Net cross-sectional area

A1 = Net cross-sectional area of

connected leg12th March, 2013


A1 = (length of leg – ½ t – nd) × t

A2 = Gross cross-sectional area of

outstanding leg A2 = (length of leg – ½ t) x t

K = constant =

Anet = A1 + A2K

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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 :

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Net effective area, i.e., Anet = A1 + A2K

Where, Anet = Net cross-sectional area

A1 = Net cross-sectional area of

connected leg(flange of the tee)

A2 = Gross cross-sectional area of

outstanding leg(web of the tee)

K = constant =

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A1 = 2 (length of leg – ½t – nd) × t

A2 = 2 (length of leg – ½ t) × t

In case of T section A1 = (bf - 2d ) × tf

A2 = Gross area of web = tw × (overall depth(h) – tf)


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. :-

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Deduction for rivet holes = number of rivets × gross diameter of one rivet × thickness of plate i.e., = n × d × tNet effective area, Anet = Gross area – Deduction for rivet holes

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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 σat = 150N/mm2

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Numerical Problems


Solution: Nominal dia. of rivet (D) =20mm Gross dia. of rivet hole (d) = 20 +1.5 = 21.5mmWidth of plat = 30 + 60 + 60+ 60 + 30 = 240mm

s = 50mm g = 60mmConsidering the failure of the plate along section line 1-2-3-4, we have; Anet1

= (b – nd) t

t = 12mm, n = 2, b = 240mm, d = 21.5mm

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Anet1 = (240 -2 × 21.5) × 12 = 3564mm2

Now for area along staggered line 1-2-5-3-6-7 Anet2 =

Where b = 240mm n = 4 d = 21.5mm n’ = 3 t = 12mm

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Anet2

= = 2073 mm2

Minimum net area of section = Anet2

Strength of the plate = 2073 × 150 = 310950 = 310.95kN

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Problem 2: Calculate the strength of a tie composed of ISA 100 × 75 × 8mm with longer legs connected by 16mm dia. rivets. Take σat = 150N/mm2

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Solution: Nominal dia. of rivet = 16mm Gross dia. of rivet = 16 + 1.5 = 17.5mmNet Area; Anet = A1 + A2K

A1 = = 628mm2

A2 = 561mm2

K= constant =

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Anet = A1 + A2K

= 628 + 561 × 0.77 = 1059.97mm2

Strength of ISA 100 × 75 × 8 = 1059.97mm2 × 150 = 158995.5N = 158.99kN

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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 σat = 150N/mm2

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Solution: Nominal dia. of rivet (D) =20mm Gross dia. of rivet hole (d) = 20 +1.5 = 21.5mm Anet = A1 + A2K

A1 = = 1592mm2

A2 = = 1136mm2

K = constant =

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K =

Net effective area Anet = A1 + A2K

= 1592 + 1136 × 0.875 = 2586mm2

Strength of tie member = 2586 × 150 = 387900N = 387.9kN

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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.

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Solution: From steel table, gross area of ISA 125 × 75 × 8mm = 15.38cm2 = 1538mm2

Net effective area =Gross area – Deduction for rivet holes Anet = 2[1538 – 21.5 × 8] = 2732mm2

Strength of tie member = 2732 × 150 = 109.8kN

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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 σat = 150N/mm2

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

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Thickness of Flange, tw = 8.4mm

Thickness of web, tf = 9mm

Overall Height = 75mm

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Case under consideration : Tee (or 2ISA) connected on same side of the gusset plate by riveting: Anet = A1 + A2K

K = constant =

Properties of ISHT 75 (From steel cables) Width of plate, b = 150mm

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Net Area of flange, A1 = (width of flange –nd) × tf

= (150 – 2 × 21.5) × 9 (Since, number of rivets(n) = 2)

= 963mm2

Gross area of web, A2 = (75 - 9) × 8.4 (Since, Area = Length of web × tw)

= 554.4mm2

K =

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Net cross sectional area, Anet = A1 + A2K

= 963 + 554.4 × 0.897 = 1460.29mm2

Tensile Strength of the T – section = Anet × σat

= 1460.29 × 150

= 219043.5N = 219.04kN

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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. σat) are known for the steel with yield stress fy.

Net sectional area required =

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i.e. Anet reqd. =

Step 2. Selection of suitable sectionTry 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)

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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 required12th March, 2013


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 (rmin ) can be obtained from steel tables.

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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.

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Step 6. Design of end connectionsThe 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 Pb and Pf)

(ii) Find the number of rivets by the relation ; Number of rivets required =

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(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.

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(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;

= τvf × l × t

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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 σat = 150N/mm2

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Solution : Data given ; Load (axial pull) = 165kN = 165 × 103 NNominal dia. of rivet, D = 18mmGross dia. of rivet, d = 18 + 1.5 = 19.5mm For PDSR, τvf = 100 N/mm2

σpf = 300 N/mm2

σat = 150 N/mm2

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Net area required = =

= 1100mm2

Assuming the gross area to be about 25% greater than the net area

Gross area = 1100 × 1.25 = 1375mm2

from steel cable, try 2 ISA gross area equal to or greater than 1375mm2

Let us try an ISA 80 × 50 × 6 mm @ 57.9 N/m = 7.46cm2 or 746mm2

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Gross area for 2 ISA = 2 × 746 = 1492mm2

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)

Anet = Gross area - Deduction for holes

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2 ISA connected back to back (with longer leg connected) on the both sides of gusset [late

Anet = 1492 – (2 × 19.5) × 6

= 1258 > 1100mm2

Hence, use 2 ISA 80 × 50 × 6 mm @ 57.9 N/m as a tension member.

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Design of riveted end connectionsShearing strength of one rivet (in double shear) = 2 × τvf ×

= 2 × 100 × = 59729.53 N ...(1) Bearing Strength of one rivet = σpf × d × t

= 300 × 19.5 × 6 = 35100 N …(2)12th March, 2013


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

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CONCLUSIONTension 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, 2013

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

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