The butt connection shows 8-22 mm diameter bolts spaced as shown below. P 50 100 50 50 100 50 ## 00 O 16 mm 00 O O 40 80 40 12 mm Steel strength and stresses are: Yield strength, Fy = 248 MPa Ultimate strength, Fu = 400 MPa Allowable tensile stress on the gross area = 148 MPa Allowable tensile stress on the net area = 200 MPa Allowable shear stress on the net area = 120 MPa Allowable bolt shear stress, Fv = 120 MPa Based on the gross area of the plate. Based on the net area of the plate. Based on block shear strength. -P -P Bolt hole diameter = 25 mm Calculate the allowable tensile load, P, under the following conditions:
The butt connection shows 8-22 mm diameter bolts spaced as shown below. P 50 100 50 50 100 50 ## 00 O 16 mm 00 O O 40 80 40 12 mm Steel strength and stresses are: Yield strength, Fy = 248 MPa Ultimate strength, Fu = 400 MPa Allowable tensile stress on the gross area = 148 MPa Allowable tensile stress on the net area = 200 MPa Allowable shear stress on the net area = 120 MPa Allowable bolt shear stress, Fv = 120 MPa Based on the gross area of the plate. Based on the net area of the plate. Based on block shear strength. -P -P Bolt hole diameter = 25 mm Calculate the allowable tensile load, P, under the following conditions:
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Kindly use the pattern for solving the problem.
Transcribed Image Text:The butt connection shows 8-22 mm diameter bolts spaced as
shown below.
P
50 100 50 50 100 50
16 mm
O
Based on the gross area of the plate.
Based on the net area of the plate.
Based on block shear strength.
40
80
40
off
O
=
12 mm
P
Steel strength and stresses are:
Yield strength, Fy = 248 MPa
Ultimate strength, Fu = 400 MPa
Allowable tensile stress on the gross area = 148 MPa
Allowable tensile stress on the net area
200 MPa
Allowable shear stress on the net area = 120 MPa
Allowable bolt shear stress, Fv = 120 MPa
Bolt hole diameter = 25 mm
Calculate the allowable tensile load, P, under the following
conditions:
P
![The butt connection shows 8-22 mm diameter bolts spaces as shown below. Steel strengths are F, = 248 MPa
and F₁ = 400 MPa. Calculate the design tensile strength based on the following:
bolt hole diameter = 25mm
Based on the gross area of the plate
Based on the net area of the plate.
Based on block shear strength.
P
P4
16 mm
16 mm
Based on the gross area of the plate.
50 100 50 50 100 50
Based on the net area of the plate.
.
Shear lag factor, W: 1.0
An
Ph 0.75 Ph 0.5 Fu Ac = 0.5 Fu UAN
Based on block shear strength.
LUL
2
>
160 (16) 2(25) (16) = 1760 mm ²
Ø Pn = 0.75 (400) (1.0) (1760)/1000 = 528 KA
.: R= 885. 12 KW
Pm 0.9Pn= 0.9 Fylg = 0.9 (248) (160) (16)/1000 = 571. 392 kW
=
Ph
+P/₂
=
*
Anr
Amr
Up=1-0; Fy= 248; Fu = 400 mpa
Am[1501.5 (25)] 16 (2) = 3600
Agv = 150 (16) (2) 4800
Ant (8025) (16) 8.80 MM ²
- mm
Agr
40
Any
3600
Agv = 4800
Ant = (80-25) (1) = 880mm ²
80
Agv = 100 (16) = 1600
-
Anb
40
Stress:
Outer Plate
6.
P/₂
w (12)
12 mm
Anv [150-15 (25)] 16 = 1800 fram ²
Agv = 150 (16) = 2400
Ant = [120 - 1.5 (25)] 16 = 1320mm ²
DERM
0₂R = 0.75Rm = 663.84KN
2
= (100- 25) (16) = 1200
2
Ant (160-2(25)) (16) = 1760 mm)
=
P
P
24W
middle plate
6 = P
P
3
W (16) 16 W
R,₁,-0.6FA+USFUAM = 12160KN)
0.6F,A+UFA 1066-240
R = 0.6FA+UFμAni= 1216.0KN)
0.6F,A+UFA 1066-240
R = 0.6FA+U₂₂FA = 960 kW
0.6F,A+UFA 885.12K
R₁ -0.6F,A+UFAL = 992.0ku
0.6F,A+UFA = 942.08k](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbb67fbec-6c43-471d-86fd-70124265fb42%2F529067e0-c282-4e73-9b4f-5100fa934436%2Foo2v4vd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The butt connection shows 8-22 mm diameter bolts spaces as shown below. Steel strengths are F, = 248 MPa
and F₁ = 400 MPa. Calculate the design tensile strength based on the following:
bolt hole diameter = 25mm
Based on the gross area of the plate
Based on the net area of the plate.
Based on block shear strength.
P
P4
16 mm
16 mm
Based on the gross area of the plate.
50 100 50 50 100 50
Based on the net area of the plate.
.
Shear lag factor, W: 1.0
An
Ph 0.75 Ph 0.5 Fu Ac = 0.5 Fu UAN
Based on block shear strength.
LUL
2
>
160 (16) 2(25) (16) = 1760 mm ²
Ø Pn = 0.75 (400) (1.0) (1760)/1000 = 528 KA
.: R= 885. 12 KW
Pm 0.9Pn= 0.9 Fylg = 0.9 (248) (160) (16)/1000 = 571. 392 kW
=
Ph
+P/₂
=
*
Anr
Amr
Up=1-0; Fy= 248; Fu = 400 mpa
Am[1501.5 (25)] 16 (2) = 3600
Agv = 150 (16) (2) 4800
Ant (8025) (16) 8.80 MM ²
- mm
Agr
40
Any
3600
Agv = 4800
Ant = (80-25) (1) = 880mm ²
80
Agv = 100 (16) = 1600
-
Anb
40
Stress:
Outer Plate
6.
P/₂
w (12)
12 mm
Anv [150-15 (25)] 16 = 1800 fram ²
Agv = 150 (16) = 2400
Ant = [120 - 1.5 (25)] 16 = 1320mm ²
DERM
0₂R = 0.75Rm = 663.84KN
2
= (100- 25) (16) = 1200
2
Ant (160-2(25)) (16) = 1760 mm)
=
P
P
24W
middle plate
6 = P
P
3
W (16) 16 W
R,₁,-0.6FA+USFUAM = 12160KN)
0.6F,A+UFA 1066-240
R = 0.6FA+UFμAni= 1216.0KN)
0.6F,A+UFA 1066-240
R = 0.6FA+U₂₂FA = 960 kW
0.6F,A+UFA 885.12K
R₁ -0.6F,A+UFAL = 992.0ku
0.6F,A+UFA = 942.08k
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