BOLTED CONNECTIONS Situation 1: The lap joint shown consists of bolts 22 mm in diameter with 25 mm diameter holes. The plates are A36 Steel with Fy 250 MPa and Fu = 400 MPa. Use xi = 50 mm, x2 = 160 mm, X3 = 60 mm and plate X3 thickness of t = 12 mm. Allowable stresses are as follows: Shear strength of bolts, Fv = 210 MPa Bearing stress on projected area, Fe = 1.5 Fu Tensile stress on gross area, Fr 0.6 Fy Tensile stress on net area, Fr = 0.5 Fu Shear on plates, Ex = 0.3 Fu 1. Determine the value of P based on bolt shear. 478.97 kN 950.40 kN a. с. 468.00 kN b. d. 504.00 kN Find the value of P based on bearing of plates. 468.00 KN 504.00 KN 2. a. 478.97 KN C. b. 950.40 kN d. 3. Compute the value of P based on tension on the gross area. 478.97 kN 468.00 kN 504.00 KN с. b. 950.40 kN d. 4. Solve the value of P based on tension on the net area. 663.66 kN 756.00 kN a. C. 823.32 kN b. d. 504.00 kN 5. Calculate the value of P based on block shear. 633.60 kN 568.80 kN 554.40 kN a. C. b. 489.60 kN d. O---

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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So 1 2 and 3
BOLTED CONNECTIONS
Situation 1:
The lap joint shown consists of bolts 22 mm in
diameter with 25 mm diameter holes. The plates
are A36 Steel with Fy = 250 MPa and Fu = 400 MPa.
Use xi = 50 mm, x2 = 160 mm, X3 = 60 mm and plate
P
thickness of f = 12 mm.
Allowable stresses are as follows:
Shear strength of bolts, Ev = 210 MPa
Bearing stress on projected area, F = 1.5 Fu
Tensile stress on gross area, Fi = 0.6 Fy
Tensile stress on net area, Fr = 0.5 Fu
Shear on plates, Fv = 0.3 Fu
1.
Determine the value of P based on bolt shear.
478.97 kN
950.40 KN
a.
C.
468.00 kN
b.
d.
504.00 kN
2.
Find the value of P based on bearing of plates.
ä.
478.97 KN
468.00 kN
b.
950.40 kN
d.
504.00 kN
3.
Compute the value of P based on tension on the gross area.
a.
478.97 kN
C.
468.00 kN
b.
950.40 kN
d.
504.00 kN
4.
Solve the value of P based on tension on the net area.
a.
663.66 kN
с.
823.32 kN
b.
756.00 kN
d.
504.00 kN
5.
Calculate the value of P based on block shear.
633.60 kN
489.60 kN
568.80 kN
554.40 kN
a.
C.
b.
d.
-----
O--
Transcribed Image Text:BOLTED CONNECTIONS Situation 1: The lap joint shown consists of bolts 22 mm in diameter with 25 mm diameter holes. The plates are A36 Steel with Fy = 250 MPa and Fu = 400 MPa. Use xi = 50 mm, x2 = 160 mm, X3 = 60 mm and plate P thickness of f = 12 mm. Allowable stresses are as follows: Shear strength of bolts, Ev = 210 MPa Bearing stress on projected area, F = 1.5 Fu Tensile stress on gross area, Fi = 0.6 Fy Tensile stress on net area, Fr = 0.5 Fu Shear on plates, Fv = 0.3 Fu 1. Determine the value of P based on bolt shear. 478.97 kN 950.40 KN a. C. 468.00 kN b. d. 504.00 kN 2. Find the value of P based on bearing of plates. ä. 478.97 KN 468.00 kN b. 950.40 kN d. 504.00 kN 3. Compute the value of P based on tension on the gross area. a. 478.97 kN C. 468.00 kN b. 950.40 kN d. 504.00 kN 4. Solve the value of P based on tension on the net area. a. 663.66 kN с. 823.32 kN b. 756.00 kN d. 504.00 kN 5. Calculate the value of P based on block shear. 633.60 kN 489.60 kN 568.80 kN 554.40 kN a. C. b. d. ----- O--
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