W = L2 = %3D 2.4 1.2 KN m L3=1.2 L1 = L3 1.2 1.2 Typical connection at all joints. m Im Li=1.2 L2=1.2 W: BC is 75 mm wide X 10 mm thick strut. 7. What is the minimum required bolt diameter (mm) at C if the allowable bolt shear stress is 68 Mpa. llin 8. Determine the stress (Mpa) in strut BC. AD is 100 9. Calculate the bolt stress (Mpa) at A if its Imm wide diameter is 20 mm. x 10 mm thick rigid bar.

W = L2 = %3D 2.4 1.2 KN m L3=1.2 L1 = L3 1.2 1.2 Typical connection at all joints. m Im Li=1.2 L2=1.2 W: BC is 75 mm wide X 10 mm thick strut. 7. What is the minimum required bolt diameter (mm) at C if the allowable bolt shear stress is 68 Mpa. llin 8. Determine the stress (Mpa) in strut BC. AD is 100 9. Calculate the bolt stress (Mpa) at A if its Imm wide diameter is 20 mm. x 10 mm thick rigid bar.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

A double-angle shape is shown in the figure. The steel is A36, and the holes are for 1/2-inch-diameter bolts. Assume the Ae = 0.75An. Section 2L5 x 3 x16 LLBB Determine the design tensile strength for LRFD. Select one: а. 78 b. 66 С. 132 d. 156
16) what is the tensile yielding capacity of flange plate of the flange connection in kips? (2 decimal places) A992: Fy=50ksi, Fu=65ksiA36: Fy=36ksi, Fu=58ksieffective bolt hole diameter = bolt hole + 1/16"beam properties: tw=0.38 in., bf=7.07 in., tf=0.63 in.Fexx=70ksi
QUESTION 8 3. check the capacity of the shear connection. A992: Fy=50ksi, Fu=65ksi A36: Fy=36ksi, Fu=58ksi effective bolt hole diameter = bolt hole + 1/16" beam properties: tw=0.295 in. Fexx=70ksi what is the shear yielding capacity of shear plate in kips? (2 decimal places)
Problem. A bolted connection shown consists of two plates 300 mm x 12 mm connected by 4 - 22 mm diameter bolts. 12 mm bolts T12 mm te12 mm P 300 75 75 75 Edge distances = 75 mm %3D dnole for tensile and rupture = dp +3 mm %3D dnole for bearing strength for Lec = d + 1.5 mm Fy = 248 MPa Fu = 400 MPa %3D En 330 MPa %3D Use LRFD design method. Determine the design strength due to tensile rupture strength of plates. (kN)
A76 x 76 x6 mm angular section shown is welded to an 8 mm thick gusset plate. The length L1 is 65 mm, L2 is 125 mm, and the cross sectional area of the angle is 929 sq.mm. Fy-248MPA and Fu -400MPA. Gusset Plate Alowable Sresse Alowable tensle stress igross ara)0.00 Alowable tereile stres net area)0SOF. Allowable shear stress (net ares)0.30F. Determine the value of P based on net area using a strength reduction coefficient of 85% O 138.24 KN O 140.14 kN O 157.93 kN O 304.00 N Next
Plate t = 8mm C230x19.9 M20 Bolts Ag = 2,530 mm? d = 229 mm T 80 tw = 5.90 mm bf = 61 mm tr = 10.5 mm F, = 345 MPa x = 15 mm F = 450 MPa 60 80 80 60 From the figure shown, which of the following most nearly gives the block shear strength? O 679,7 KN O 1.013.8 KN O 872.9 KN 691.4 KN
sted ingid 250MM B k steel bar bolt AUR rigid bar 12 mm thick 8 mm Brass 350mm Jointd 16 mene thick 350mm JointB 250 mm Xg = 20x10% Eg = 90 G Pa -6 A, = 12x10/0 Es = 200 G Pa thickness> 16 mm 8 mm bolt %3D 12 mmi thick shear strength f bolf bearing streng th of holt= 100 la 50 MPa %3D Assume no failure will take place in steel or brass. Temperature chauge brass steel on Temperature change on Determine thai can be applied to system - the max
10) what is the bolt shear capacity of the flange connection in kips? (2 decimal places). Given:A992: Fy=50ksi, Fu=65ksiA36: Fy=36ksi, Fu=58ksieffective bolt hole diameter = bolt hole + 1/16"beam properties: tw=0.38 in., bf=7.07 in., tf=0.63 in.Fexx=70ksi
Material Strengths: Fy = 248 MPa, F = 400 MPa For the two lines of bolt holes, the pitch is the value that will give a net area DEFG equal to the one along ABC. The diameter of the holes is 22 mm. The thickness of the plate is 12 mm. Determine the LRFD design tensile strength based on yielding on gross area. D A to 50 1 ÓB 50 50 ic O 410 kN O 464 kN O 446 kN O 401 KN F IG Pitch Pitch KISS|Y
A 16 mm thick tension member is connected by two 6.25 mm spliced plates as shown. The tension member carries a service loads of dead load of 110 kN and a live load of 100 kN. 40, 80 , 40 , 40, 80 40 1625 mm 16 mm 1625 mm Fy 248 MPa Fu = 400 MPa Diam. of bolts = 16 mm Fnv = 300 MPa O Determine the nominal strength for one bolt due to shear. O Determine the nominal strength for one bolt due to bearing strength of the connection. ® Determine the number of bolts required for the connection.
The given plate below with width of 200 mm andthickness of 16 mm is to be connected to two plates ofthe same width with half the thickness by 20 mmdiameter rivets as shown. The rivet hole is 2 mm greaterthan the rivet diameter. Allowable tensile stress on netarea is 0.6Fy. Allowable bearing stress is 1.35Fy. plate: fy = 241 mpa fu = 4141 mpa rivet: fv = 152 mpa a. Determine maximum load P without exceeding allowable tensile stress on plate b. Determine maximum load P without exceeding allowable shear stress on rivets c. Determine maximum load P without exceeding allowable bearing stress between plates and rivets
Six rivets are used in the connection shown in the figure. If both P=50KN, what is the minimum diameter rivet is necessary so as not exceed the 70MPa allow shearing stress? What thickness of plate is required so as not exceed a bearing stress of 140MPA? p= 50 80 mm 80 mm 100 mm