The figure shows a plate having a width of 400 mm and thickness of 12 mm is to be 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. 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 (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 - (mm²) (9) Determine the net area of the (14) Compute the allowable strength of the connection due to fracture in the net section. (ASD Method) – (kN) (15) Computethe design strength of the connection due to yielding of the gross section. (LRFD Method) - (kN) path 1-2-3 if b = 30 m - (mm2) (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 the connection due to shear of failure of bolts. Fy = 270 MPa - (kN) Computethe allowable strength in the chain, the quantity where S is the longitudinal center to center (pitch) of (16) Compute the design any consecutive holes, in millimeters and g is the transverse center to center spacing (gage) between fastener gage lines in millimeters. If a = N mm, strength of the connection due to fracture in the net section. c = M mm and d = L mm. (LRFD Method) – (kN) (17) Which of the following 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) N = 80 + B (13) of the connection due to yielding of the gross section. (ASD Method) – (kN) P. M = 100 + Y L= 100 + B la b
The figure shows a plate having a width of 400 mm and thickness of 12 mm is to be 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. 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 (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 - (mm²) (9) Determine the net area of the (14) Compute the allowable strength of the connection due to fracture in the net section. (ASD Method) – (kN) (15) Computethe design strength of the connection due to yielding of the gross section. (LRFD Method) - (kN) path 1-2-3 if b = 30 m - (mm2) (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 the connection due to shear of failure of bolts. Fy = 270 MPa - (kN) Computethe allowable strength in the chain, the quantity where S is the longitudinal center to center (pitch) of (16) Compute the design any consecutive holes, in millimeters and g is the transverse center to center spacing (gage) between fastener gage lines in millimeters. If a = N mm, strength of the connection due to fracture in the net section. c = M mm and d = L mm. (LRFD Method) – (kN) (17) Which of the following 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) N = 80 + B (13) of the connection due to yielding of the gross section. (ASD Method) – (kN) P. M = 100 + Y L= 100 + B la b
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
Please answer nos. 10, 11 and 12
N=82
M=132
L=102
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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