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 . 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 39 ✪ 60:60 200 mm 30 KN 20 mm o bolts

Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
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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 3, 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 3, 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) = 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 =
=
fone 2 =
= 49.857
Stress
=
Rv₁ =
= RHI=0
RV₂
Forme (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
= 7.5 KN (vertical force 1)
mx
му
my
6000 kn.mum
6000x60
44100
=81632
(safe)
= 18. 693802 KN
210-204
= 8.16 32 35
= 7.5
60007 75
44100
kn
6000x
44100
(10.2040) + (7.5 + 2. 163225) ² Kn
4582
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) = 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 = = fone 2 = = 49.857 Stress = Rv₁ = = RHI=0 RV₂ Forme (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 = 7.5 KN (vertical force 1) mx му my 6000 kn.mum 6000x60 44100 =81632 (safe) = 18. 693802 KN 210-204 = 8.16 32 35 = 7.5 60007 75 44100 kn 6000x 44100 (10.2040) + (7.5 + 2. 163225) ² Kn 4582 18 693202 X (20) 2 = 59 5044 69 1000 тра KN (safe) Kn
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