Determine the maximum load P that the frame can withstand without bending the A992 BC steel part. Due to the split ends of the element, consider that the B and C supports act as joints for buckling on the x - x axis and as fixed supports for buckling on the y-y axis. Note : A similar problem is already solved on bartl*by.i have attached its screen shot here. You can refer it. 3:27 PM | 47.8KB/swww.bartleby.com/questions-and-ansv CQ SEARCH ASK CHAT VX MATH SOLV0.9 mK17-37. Determine the greatest load P the frame willsupport without causing the A-36 steel member BC tobuckle. Due to the forked ends on the member, considerthe supports at B and C to act as pins for x-x axis bucklingand as fixed supports for y-y axis buckling.CxExpert Solution-1.2 m-B4G75 mmJ-1.255 الله الله25 mmTranscribed Image Text: 17-37. Determine the greatest loadP the frame will support without causing the A-36 steelmember BC to buckle. Due to the forked ends on the member,consider the supports at B and C to act as pins for x-x axisbuckling and as fixed supports for y-y axis buckling. -1.2 m--1.2 m- 0.9 m 75 mm CT 25 mmgo3:29 PM | 42.8KB/sSEARCH个Expert Solution←www.bartleby.com/questions-and-ansv CASKFBCLet,Step 1Drawing the FBD of the given figure:0.91.51.2 mF1.5BCCHAT √x MATH SOLV1.20.94GStep 2Taking moment about A equal to zero:ΣM₁ = 0 (ccw+ve)A55 ) الله الله1.2 m1.2 - P(2.4) = 0P=0.3FBC ...(1)E = young's modulus of the frame A-36 steel= 200 x 10³ MPa (from PSG data book)4K = Length fixity coefficient = 1 (both endspinned)3:29 PM | 48.9KB/sSEARCH个www.bartleby.com/questions-and-ansvK = Length fixity coefficient = 1 (both endspinned)PL = length of the member BC = 1.5 m = 1500mmStep 3Now calculating critical load (Pcrx) for x-xbuckling:crxP=cryASK 即CHAT √x MATH SOLV2 EI=π᾽ΕΙ(KL)²π²×200×10³ ×= 771062.8 N4GNow calculating critical load (Pcry) for y-ybuckling:(1×1500)²π᾽ΕΙ(KL)²●●●25×75³1235525³ ×7543:29 PM | 12.6KB/s个←Q SEARCH ASK CHAT √x MATH SOLVwww.bartleby.com/questions-and-ansv Ccry(KL)²π² ×200×10³ ×2= 85673.6 N(1x1500)²Therefore, Pcry = FBCHence from eq. 1 we get:P=0.3F55 الله الله4GStep 4Hence for safe condition consideringminimum critical load Pcry equal to theload on the member BC.25³ ×7512BC= 0.3x85673.6= 25702.08 N= 25.7 kN●●●Hence the required greatest load P that theframe will support without causing the 25 mm35 mmm1mm

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Answered: Determine the maximum load P that the…

Elements Of Electromagnetics

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Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Determine the maximum load P that the frame can withstand without bending the A992 BC steel part. Due to the split ends of the element, consider that the B and C supports act as joints for buckling on the x - x axis and as fixed supports for buckling on the y-y axis.

Note : A similar problem is already solved on bartl*by.i have attached its screen shot here. You can refer it.

3:27 PM | 47.8KB/s
www.bartleby.com/questions-and-ansv C
Q SEARCH ASK CHAT VX MATH SOLV
0.9 m
K
17-37. Determine the greatest load P the frame will
support without causing the A-36 steel member BC to
buckle. Due to the forked ends on the member, consider
the supports at B and C to act as pins for x-x axis buckling
and as fixed supports for y-y axis buckling.
Cx
Expert Solution
-1.2 m-
B
4G
75 mm
J
-1.2
55 الله الله
25 mm
Transcribed Image Text: 17-37. Determine the greatest load
P the frame will support without causing the A-36 steel
member BC to buckle. Due to the forked ends on the member,
consider the supports at B and C to act as pins for x-x axis
buckling and as fixed supports for y-y axis buckling. -1.2 m-
-1.2 m- 0.9 m 75 mm CT 25 mm
go
3:29 PM | 42.8KB/s
SEARCH
个
Expert Solution
←
www.bartleby.com/questions-and-ansv C
ASK
FBC
Let,
Step 1
Drawing the FBD of the given figure:
0.9
1.5
1.2 m
F
1.5
BC
CHAT √x MATH SOLV
1.2
0.9
4G
Step 2
Taking moment about A equal to zero:
ΣM₁ = 0 (ccw+ve)
A
55 ) الله الله
1.2 m
1.2 - P(2.4) = 0
P=0.3FBC ...(1)
E = young's modulus of the frame A-36 steel
= 200 x 10³ MPa (from PSG data book)
4
K = Length fixity coefficient = 1 (both ends
pinned)
3:29 PM | 48.9KB/s
SEARCH
个
www.bartleby.com/questions-and-ansv
K = Length fixity coefficient = 1 (both ends
pinned)
P
L = length of the member BC = 1.5 m = 1500
mm
Step 3
Now calculating critical load (Pcrx) for x-x
buckling:
crx
P
=
cry
ASK 即CHAT √x MATH SOLV
2 EI
=
π᾽ΕΙ
(KL)²
π²×200×10³ ×
= 771062.8 N
4G
Now calculating critical load (Pcry) for y-y
buckling:
(1×1500)²
π᾽ΕΙ
(KL)²
●●●
25×75³
12
3
55
25³ ×75
4
3:29 PM | 12.6KB/s
个
←
Q SEARCH ASK CHAT √x MATH SOLV
www.bartleby.com/questions-and-ansv C
cry
(KL)²
π² ×200×10³ ×
2
= 85673.6 N
(1x1500)²
Therefore, Pcry = FBC
Hence from eq. 1 we get:
P=0.3F
55 الله الله
4G
Step 4
Hence for safe condition considering
minimum critical load Pcry equal to the
load on the member BC.
25³ ×75
12
BC
= 0.3x85673.6
= 25702.08 N
= 25.7 kN
●●●
Hence the required greatest load P that the
frame will support without causing the

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ISBN:

9780190698614

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Sadiku, Matthew N. O.

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