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Determine the maximum load P that the frame can withstand without bending

Determine the maximum load P that the frame can withstand without bending

Elements Of Electromagnetics
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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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.

 

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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
Transcribed Image Text: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
25 mm 35 mm m 1m m
Transcribed Image Text:25 mm 35 mm m 1m m
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