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Please answer nos. 16 and 17 N=82 M=132 L=102

Please answer nos. 16 and 17 N=82 M=132 L=102

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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Please answer nos. 16 and 17 N=82 M=132 L=102
(7) Determine the net area of the path 1-2-3-4 if b = 30 mm - (mm²) (8) Determine the net area of the path 1-2-4 if b = 30 mm – (mm?) (9) Determine the net area of the path 1-2-3 if b = 30 m - (mm?) (10) Determine the net area of the path 2-3 if b = 30 mm - (mm²) (11) Determine the net area of the path 2-4 if b = 30 mm - (mm²) (12) Compute the tensilestrength of (17) Which of the following The figure shows a plate having a width of 400 mm and thickness of connected to another plate by 34 mm o bolts as shown in the figure 2.22. Assume diameter of holes to be 2 mm larger than the diameter of the bolts. Use A 36 steel plate with yield strength F, = 248 MPa and minimum tensile strength F, = 400 MPa. mm is to be (14) Compute the allowable strength of the connection due to fracture in the net section. (ASD Method) – (kN) (15) For a chain of holes extending across a part in any diagonal or zigzag line, the net width of the part shall be obtained by deducting from the gross width the sum of the diameters or slot dimensions of all holes in the chain and adding, for each gage space Compute the design strength of the connection due to yielding of the gross section. (LRFD Method) – (kN) (16) in the chain, the quantity , where S is the longitudinal center to center (pitch) of Compute the design strength of the connection due to fracture in the net section. any consecutive holes, in millimeters and g is the transverse center to center spacing (gage) between fastener gage lines in millimeters. If a L mm. mm, c = M mm and d = (LRFD Method) – (kN) the connection due to shear of failure of bolts. Fy = 270 MPa gives the nearest value of "b" so that the net width along bolts 1- 2-3-4 is equal to the net width along bolts 1-2-4?- (mm²) - (kN) (13) N = 80 + B M = 100 + Y Computethe allowable strength of the connection due to yielding of the gross section. (ASD Method) – (kN) L= 100 + B la b P NE ---
Transcribed Image Text:(7) Determine the net area of the path 1-2-3-4 if b = 30 mm - (mm²) (8) Determine the net area of the path 1-2-4 if b = 30 mm – (mm?) (9) Determine the net area of the path 1-2-3 if b = 30 m - (mm?) (10) Determine the net area of the path 2-3 if b = 30 mm - (mm²) (11) Determine the net area of the path 2-4 if b = 30 mm - (mm²) (12) Compute the tensilestrength of (17) Which of the following The figure shows a plate having a width of 400 mm and thickness of connected to another plate by 34 mm o bolts as shown in the figure 2.22. Assume diameter of holes to be 2 mm larger than the diameter of the bolts. Use A 36 steel plate with yield strength F, = 248 MPa and minimum tensile strength F, = 400 MPa. mm is to be (14) Compute the allowable strength of the connection due to fracture in the net section. (ASD Method) – (kN) (15) For a chain of holes extending across a part in any diagonal or zigzag line, the net width of the part shall be obtained by deducting from the gross width the sum of the diameters or slot dimensions of all holes in the chain and adding, for each gage space Compute the design strength of the connection due to yielding of the gross section. (LRFD Method) – (kN) (16) in the chain, the quantity , where S is the longitudinal center to center (pitch) of Compute the design strength of the connection due to fracture in the net section. any consecutive holes, in millimeters and g is the transverse center to center spacing (gage) between fastener gage lines in millimeters. If a L mm. mm, c = M mm and d = (LRFD Method) – (kN) the connection due to shear of failure of bolts. Fy = 270 MPa gives the nearest value of "b" so that the net width along bolts 1- 2-3-4 is equal to the net width along bolts 1-2-4?- (mm²) - (kN) (13) N = 80 + B M = 100 + Y Computethe allowable strength of the connection due to yielding of the gross section. (ASD Method) – (kN) L= 100 + B la b P NE ---
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