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6.9 Problem: A bracket shown in the figure 6.9 carries an eccentric load of 30 &N at an eccentricity of 200 mm. Diameter of bolts is 20 mm a. Use the reduced eccentricity method of analysis. Compute the shearing stress of bolt A. Compute the shearing stress of bolt B. 3 Compute the shearing stress of bolt C. 75 75 200 mm 30 KN

6.9 Problem: A bracket shown in the figure 6.9 carries an eccentric load of 30 &N at an eccentricity of 200 mm. Diameter of bolts is 20 mm a. Use the reduced eccentricity method of analysis. Compute the shearing stress of bolt A. Compute the shearing stress of bolt B. 3 Compute the shearing stress of bolt C. 75 75 200 mm 30 KN

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Subject: Steel design- Bolted Steel Connection - BOLTS SUBJECTED TO ECCENTRIC LOADS

 

 

QUESTIONS:

HOW MANY BOLTS (N) SHOULD APPLY FOR THIS PROBLEM IS IT 6 or 4? Please justify your answers and explain and please refer the green circle in the picture. Kindly correct what N should use if 6 or 4. Thank you

 

 

6.9 Problem: A bracket shown in the figure 6.9 carries an eccentric load of 30 kN at an eccentricity of 200 mm. Diameter of bolts is 20 mm Ø. Use the reduced eccentricity method of analysis M = .. Compute the shearing stress of bolt A. Compute the shearing stress of bolt B. 3 Compute the shearing stress of bolt C. Solution : E centric Shearing stress of Z= = (x² + y²) 2 = eccentricity Diameter of I moment = Pxe 441 00 mm 2 RV₁ = load = 30 KN vertical load 2 = = ; . Fv = = 200 mm = 30x 200 = 6000 30 4 Horizontal Load = O 75 75 bolts bolt A 4x (25²) + 6x (60²) 4 7.5 kn RV2= Horizontal load ₂ = RH₂ = = 20mm PO Mx 60 60 000 6000 × 60 44100 = 8. 163235 kn 18. 69388 Kn my 2 (vertical load) = Resultat Force=√(10.2040)² + (7.5+ 8·1632)² Fr 1x (20)² = 18·693x 1000 18. 693 x 1000 x (20)² 200mm kn.mm 6000 x 75 44100 = 10.201082 KN 20 mm & bolts 30 KN =59.5044 тра (safe) 75 75 Ⓒ veretical 60:60 verstical 200 mm 199 Horizontal Horizontal .. Resultat ... Resultant Shearing Stress of Bolt Z= = (x² + y² 2 = 44100 mm² M = Pxe sheating - Horizontal Fone 2 = 20 mm o bolts vertical Force 2 Horizontal force 1 Shearling shxss z = M= Pxe Fr= Ax fv = R 30 KN 44100 mm 2 Fine Force 2 = = fire 2 = = 49.857 Stress = Rv₁ = = RHI=0 RV₂ Forle (B) = √ (7-5+ 8.16323) ² = 15.6632 κη 15-66 32 x 1000 퓨가 (20)2 R.V2 30 x 200 RH2 RH₂ 349. RHI (fr) = bolt тра = B = 2 mx = 7.5 KN (vertical force 1) му my 6000kn.mum 4582 6000x60 44100 = 8 16 32 (safe) 210-204 = 18. 693802 KN = 8.16 32 35 60007 75 44100 = 7.5 kn 6000x 44100 (10.2040) ²+ (7.5 + 2. 163225) ² Kn 18 693202 X (20) 2 = 59 5044 69 1000 тра KN (safe) Kn
Transcribed Image Text:6.9 Problem: A bracket shown in the figure 6.9 carries an eccentric load of 30 kN at an eccentricity of 200 mm. Diameter of bolts is 20 mm Ø. Use the reduced eccentricity method of analysis M = .. Compute the shearing stress of bolt A. Compute the shearing stress of bolt B. 3 Compute the shearing stress of bolt C. Solution : E centric Shearing stress of Z= = (x² + y²) 2 = eccentricity Diameter of I moment = Pxe 441 00 mm 2 RV₁ = load = 30 KN vertical load 2 = = ; . Fv = = 200 mm = 30x 200 = 6000 30 4 Horizontal Load = O 75 75 bolts bolt A 4x (25²) + 6x (60²) 4 7.5 kn RV2= Horizontal load ₂ = RH₂ = = 20mm PO Mx 60 60 000 6000 × 60 44100 = 8. 163235 kn 18. 69388 Kn my 2 (vertical load) = Resultat Force=√(10.2040)² + (7.5+ 8·1632)² Fr 1x (20)² = 18·693x 1000 18. 693 x 1000 x (20)² 200mm kn.mm 6000 x 75 44100 = 10.201082 KN 20 mm & bolts 30 KN =59.5044 тра (safe) 75 75 Ⓒ veretical 60:60 verstical 200 mm 199 Horizontal Horizontal .. Resultat ... Resultant Shearing Stress of Bolt Z= = (x² + y² 2 = 44100 mm² M = Pxe sheating - Horizontal Fone 2 = 20 mm o bolts vertical Force 2 Horizontal force 1 Shearling shxss z = M= Pxe Fr= Ax fv = R 30 KN 44100 mm 2 Fine Force 2 = = fire 2 = = 49.857 Stress = Rv₁ = = RHI=0 RV₂ Forle (B) = √ (7-5+ 8.16323) ² = 15.6632 κη 15-66 32 x 1000 퓨가 (20)2 R.V2 30 x 200 RH2 RH₂ 349. RHI (fr) = bolt тра = B = 2 mx = 7.5 KN (vertical force 1) му my 6000kn.mum 4582 6000x60 44100 = 8 16 32 (safe) 210-204 = 18. 693802 KN = 8.16 32 35 60007 75 44100 = 7.5 kn 6000x 44100 (10.2040) ²+ (7.5 + 2. 163225) ² Kn 18 693202 X (20) 2 = 59 5044 69 1000 тра KN (safe) Kn
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