Summary of Lessons 1-16, Steel Reinforcement PDF

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WELDED BRACKET CONNECTION (SHEAR AND BENDING) From the bending stress and shearing stress shown in the figure, the maximum bending stress and maximum shearing stress do not occur at the same point on the weld. Maximum bending stress occurs at the top and bottom while the maximum shearing stress occu...

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WELDED BRACKET CONNECTION (SHEAR AND BENDING)

From the bending stress and shearing stress shown in the figure, the maximum bending stress and maximum shearing stress do not occur at the same point on the weld. Maximum bending stress occurs at the top and bottom while the maximum shearing stress occurs at the mid-point of the weld. So we cannot combine the two max stresses at any point, but in actual practice we could assume a uniform variation of shearing stress and therefore we may combined vectorially the maximum values of the two stresses.

Consider 1mm width of the fillet weld fx=

MC I

;

Where M=Pe L C= 2

( 1 ) L3 I= 12 fv=

P L(1)

Resultant stresses: R=

√ (fx)2 +( fv )2

R=0.707 t (0.3) Fu Where: t= size of fillet welds Fu= min. tensile strength of fillet weld

SAMPLE PROBLEM: 1. A bracket shown in figure is welded by a fillet weld to a column section. It carries an eccentric load of 80kN acting at 200mm from the fillet weld shown. A) Determine the max. bending stress on the fillet weld. B) Using elastic method, determine the maximum force per mm to be resisted by the fillet weld shown. C) Determine the size of fillet weld to carry the load using E70 electrodes with Fu=485MPa .

SOLUTIONS: a) Max bending stress per mm of fillet weld:

fv=

MC I

M = 80000(200) 6 M = 16 ×10 N.mm

250 2

C=

C = 125

I=

250 ¿ ¿ ¿3 1¿ ¿

I = 1302083.33

fb=

MC I

fb=

(125)16 × 106 1302083.33

fb=1536 N . mm

b) Max force per mm to be resisted by the fillet weld: fv =

V A

fv =

80000 1(250)

fv = 320

R=

N mm

√(320)2+(1536 )2

R = 1569 N/mm

c) Size of fillet weld to carry the load: t=

1569 0.707 (1)( 0.3 )( 485 )

t = 15.25mm say 16mm

2) From the given welded connection shown in figure, an eccentric load of 455kN is acting at the distance of 150mm from the flange of the column section. An E70 fillet weld is used on both sides of the bracket with a minimum tensile strength of Fu=485MPa.

a) Compute the bending stress per mm resisted by the fillet weld b) Compute the maximum force per mm to be resisted by the fillet weld. c) Compute the size of the fillet weld required for the connection shown.

SOLUTIONS: a) Bending stress per mm resisted by the fillet weld. Consider 1mm width of fillet weld: I = (1) (250)3(2) / 12 I = 2604166.67 M = 45000(150) M = 6750000 N.mm C = 125 fx = MC / I fx = 6750000(125) / 2604166.67 fx = 324 N/mm

b) Maximum force per mm to be resisted by the fillet weld: P/L = 45000/2(230) P/L = 90 R2 = (323)2 + (90)2 R = 335.30 N/mm

c) Size of fillet weld using E70 electrodes: 335.30 = 0.707 t (1) (0.3) (485) t = 3.26 say 4 mm use 4mm fillet weld

3) A bracket shown in figure is subjected to an eccentric load of 178 kN acting at 100 mm from the face of the flange of the column section. The angular section is welded to the flange of the column section by an E70 fillet weld with Fu=485 MPa.

a) Compute the moment of inertia of the fillet welds. b) Compute the maximum force per

mm to be resisted by the fillet weld neglecting the shear force to be carried by end returns. c) Compute the size of fillet weld

to carry such load.

SOLUTIONS: a) Moment of inertia of fillet welds: Consider 1 mm width of weld. A1 = 1(250)(2) A1 = 500 A2 = 12(1)(2) A2 = 24 A = A1 + A2 A = 500 + 24 A = 524

Aȳ = A1y1 + A2y2 524ȳ = 500(125) + 24(0) ȳ = 119.27 mm Ix = (1(250)3 / 12) (2) + 250(1)(2)(5.73)2 + 12(1)(2)(119.27)2 Ix = 2.962 x 106 mm4

b) Maximum force per mm that the fillet weld can resist. Neglect the direct shear to be carried by the end returns. Due to bending: fx = MC / I M = 178000(100) M = 17.8 x 106 N.mm fx = 17.8 x 106 (130.73) / 2.962 x 106 fx = 785.6 N/mm Due to direct shear: fv = P / L fv = 178000/2(250) fv = 356 N/mm R2 = (356)2 + (785.6)2 R = 862.50 N/mm

c) Size of fillet weld. 862.50 = 0.707 t (1)(0.3)(485) t = 8.38 mm

use t = 8.5 mm fillet weld

WELDED BRACKET CONNECTION (SHEAR AND TORSION)

Eccentrically loaded connections as shown in the figure, is subjected to shear and torsion.

Torsional Moment: M = Pe M = P (105+200) M = 305 P

Torsional Stressses: fx = M y / J fy = M x / J

Polar moment of inertia: J = Ix + Iy

Consider 1mm strip of weld: Ix = (1)(200)3/12 + 150(1)(100)2(2) Ix = 3.67 x 106 mm4

Iy = [ (1)(150)3/12 + 1(150)(30)2 ] 2 + 200(1)(45)2 Iy = 1.24 x 106 mm4

J = Ix +Iy

J = 3.67x106 + 1.24x106 J = 4.91x106

Direct shear stress: fv = P / L fv = P / 150+150+200 fv = P / 500 N/mm

R2 =

√(fx)2 +(fy + fv )2

R = max force per mm to be resisted by the fillet weld R = 0.707 t (1)(0.3) Fu t = size of fillet weld required Fu = min tensile strength of fillet weld

SAMPLE PROBLEM: 1) A bracket shown in the figure is carrying an eccentric load of 80 kN. Use an E70 electrodes with Fu= 485MPa. a) Determine the value of the polar moment of inertia. b) Determine the maximum force per mm to be resisted by the fillet weld. c) Determine the size of the E70xx fillet weld required for the bracket.

SOLUTIONS: a) Polar moment of inertia: Consider 1 mm of weld. J = (1)(250)3(2) / 12 + 250(1)(2)(100)2 J = 7604166.67 mm4 Check using the formula: J = ∑ L ( L2/12 + x2 + y2) J = 250(2) [ (250)2/12 + (100)2 + (0)2 ] J = 7604166.67 mm4

b) Maximum force per mm to be resisted by the fillet weld: Torsional moment:

M = 80000(200) M = 16000000 N.mm M = 16 x 106 N.mm

fx = M y / J fx = 16 x 106 (125) / 7604166.67 fx = 263 N/mm fy = M x / J fy = 16 x 106 (100) / 7604166.67 fy = 210.4 N/mm P/L = 80000/250(2) P/L = 160 N/mm R2 = (263)2 + (370.4)2 R = 454.27 N/mm

c) Size of E70 electrodes fillet weld required for the bracket: 454.27 = 0.707 t (1)(0.3)(485) t = 4.42 use t = 5mm fillet weld

2) The welded bracket connection is subjected to an eccentric load of 45kN acting as shown in the figure. An E70 fillet weld is used with a minimum tensile strength Fu=485 MPa. a) Compute the centroid of the fillet welds measured from BC. b) Compute the moment of inertia at the centroid of the group of fillet welds. c) Compute the appropriate size of the fillet welds to be used.

SOLUTIONS: a) Centroid of fillet: L = L1 + L2 + L3 L = 150 + 200 + 150 L = 500 Lx = L1x1 + L2x2 + L3x3 500x = 150(75) + 200(0) + 150(75) x = 45mm from BC

b) Polar moment of inertia: Consider 1 mm width of inertia:

2 Ix = [150(1)(100) ] 2 + (1)(200)3 / 12 Ix = 3.67 x 106

Iy = [ 1(150)3/12 + 1(150)(30)2 ] 2 + 200(1)(45)2 Iy = 1.24 x 106 J = Ix + Iy J = 3.67 x 106 + 1.24 x 106 J = 4.91 x 106 mm4

c) Size of fillet welds: M = 45000(200 + 105) M = 13725000 N.mm fx = MC / J fx = 13725000(100) / 4.91 x 106 fx = 276.53 N/mm fy = MC / J fy = 13725000(105) / 4.91 x 106 fy = 293.51 N/mm fv = P / L fv = 45000 / 500 fv = 90 N/mm Max force per mm to be resisted by the fillet weld: R2 = (383.51)2 + (279.53)2 R = 474.57 N/mm

Size of fillet weld: R = 0.707 t (1)(0.3) Fu 474.57 = 0.707 t (1)(0.3)(485) t = 4.61 mm say 5mm use 5mm fillet weld

3) The bracket shown in figure is connected to its support with a 6 mm fillet weld using E60 electrodes Fu=415 MPa. a) Compute the torsional moment to be resisted by the fillet weld b) Compute the location of the instantaneous center of rotation measured from the center of the fillet welds. c) Compute the maximum stress in the fillet welds.

SOLUTIONS:

SOLUTIONS: a) Torsional moment to be resisted by fillet welds.

Locate centroid of fillet welds. L = L1 + L2 + L3 L = 100 + 300 + 100 L = 500 mm L x = L1 x1 + L2 x2 + L3 x3 500 x = 100(50) + 300(0) + 100(50) x = 20 e = 250 – x e = 250 – 20 e = 230 mm M = Pe M = 60000(230) M = 13.8 x 106 N.mm M = 13.8 kN.m

b) Location of instantaneous center of rotation: Polar moment of inertia of group of fillet welds. Consider 1 mm strip. Ix = (1)(300)3 / 12 + (1)(100)(150)2(2) Ix = 6750000 mm4 Iy = [ (1)(20)3 / 3 + (1)(80)3 / 3 ] 2 + 300(1)(20)2 Iy = 466667 mm 4 J = Ix + Iy J = 6750000 + 466667 J = 7216667 h = J / eL h = 7216667 / 230(500) h = 62.75 mm

c) Maximum stress in the fillet welds: Maximum force per mm to be resisted by the fillet welds. d=

√(142.75 )2+(150 )2

d = 207.07 mm R=Md/J R = 60000(230)(207.07) / 7216667 R = 395.97 N/mm

Max stress in the fillet weld: R = 0.707 t (1)Fv 395.97 = 0.707(6)(1) Fv Fv = 93.34 MPa (max stress in the fillet weld)

WELDED BEAM CONNECTIONS

SAMPLE PROBLEM:

1) A W 21 x 68 beam is connected to a column section by a welded connection using a 75mm x 75mm x 10mm as its web angle welded to the flanges of the column section as shown in the figure. The fillet weld is an E60 electrodes having a nominal tensile strength of Fu = 415MPa. The beam supports reaction of 196 kN. The shop and field welds is 7mm in size for both column and beam web connection. Use A 36 steel with Fy = 248 MPa. a) Investigate if the shop weld to beam connection is adequate. b) Investigate if the field weld to column flange connection is adequate, if not what size of fillet weld is required. c) Investigate if the welded web angle thickness is adequate.

a) Adequacy of shop weld to beam connection.

Consider 1mm thick of fillet weld Locate centroid of fillet welds. L = L1 + L2 + L3 L = 65 + 220 + 65 L = 350 Lx = L1x1 + L2x2 + L3x3 350x = 65(32.5) + 220(0) + 65(32.5)

x = 12.07 mm

Polar moment of inertia of fillet welds: Ix = (1)(220)3/12 + 65(1)(110)2(2) Ix = 2460333.33 Iy = (1)(52.93)3(2)/3 + (1)(12.07)3(2)/3 + (1)(220)(12.07)2 Iy = 132081.55 J = Ix + Iy J = 2460333.33 + 132081.55 J = 2592414.88

Torsional moment: M = 98000(62.93) M = 6167140 N.mm Torsional stress: fx = M y / J fx = 6167140(110) / 2592414.88 fx = 261.68 N/mm fy = M x / J fy = 6167140(52.93) / 2592414.88 fy = 125.92 N/mm

Direct shear stress: fv = R / L fv = 98000 / 350 fv = 280 N/mm

Maximum force per mm to be resisted by fillet weld: RA2 = (405.92)2 + (261.68)2 RA = 482.96 N/mm Shear capacity of the 7mm fillet weld: q = 0.707 t (1)(0.3) Fu q = 0.707 (7)(1)(0.3)(415) q = 616.15 N/mm > 482.96 N/mm The 7 mm shop fillet weld to the beam web is adequate.

b) Adequacy of the fillet weld to column flange:

NSCP Specs and AISC Specs. The field welds are subjected to a rotation effect which causes the web angles to be forced against the beam web at the top and pushed apart at the bottom, tending to shear horizontally the fillet weld. The usual practice is to consider that the neutral axis dividing the tension from compression is located 1/6 of the way down from the top of the angles. The horizontal shear stress is assumed to vary from zero at point A, to a maximum at the bottom of the angles.

∑MA = 0 F h = 98000(75) F(2/3)(183.33) = 98000(75) F = 60137.5 N F = fx (183.33) / 2 60137.5 = fx (183.33) / 2 fx = 656.06 N/mm

Direct shear stress: fv = 98000 / 220 fv = 445.45 N/mm

Maximum force per mm to be resisted by the fillet weld: R2 = (445.45)2 + (656.06)2 R = 793 N/mm

Capacity of the 7mm fillet welds:

q = 0.707(7)(1)(0.3)(415) q = 616.15 N/mm < 793 N/mm (not adequate)

Size of fillet weld required for the column flange weld: R = 0.707 t (1)(0.3) Fu 793 = 0.707 t (1)(0.3)(415) t = 9 mm use 9 mm fillet weld

c) Adequacy of web angle: q = 793 N/mm (Capacity of 9 mm fillet weld on the column flange) For the length of 220 mm fillet weld: Capacity of fillet weld = Capacity of angle due to shear 793(220) = tw (220)(0.40)Fy 793 = tw (0.40)(248) tw = 8 mm required thickness

The angular section is adequate has A thickness of 10mm > 8mm (required)

2) Two framing angles are welded to the web of a wide flange using A36 steel (Fy = 250 MPa) and E70 electrodes Fu = 485 MPa. Using elastic method and SMAW process

a) Determine the polar moment of inertia of the group of welds. b) Determine the fillet weld size of the connection shown. c) Determine the thickness of the angle.

SOLUTIONS: a) Polar moment of inertia of the group of welds A1 = 62.5(1)(2) A1 = 125 A2 = 300(1) A2 = 300

A A1 + A2 A = 125 + 300 A = 425 mm...