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Required information Deformable rods BD and CF are attached to a rigid rod ABC as shown. Rod BD is made of steel (Est-200 GPa) and has an uniform 23x3.8 mm² cross section. Rod CF is made of aluminum (Ea=70 GPa) and has an uniform 31x3.5 mm² cross section. If a load P=19 KN were applied at A, determine the deflection at A. A 125 mm D B 225 mm- 225 mm 150 mm E

Required information Deformable rods BD and CF are attached to a rigid rod ABC as shown. Rod BD is made of steel (Est-200 GPa) and has an uniform 23x3.8 mm² cross section. Rod CF is made of aluminum (Ea=70 GPa) and has an uniform 31x3.5 mm² cross section. If a load P=19 KN were applied at A, determine the deflection at A. A 125 mm D B 225 mm- 225 mm 150 mm E

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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determine deflection at A,

Required information Deformable rods BD and CF are attached to a rigid rod ABC as shown. Rod BD is made of steel (Est-200 GPa) and has an uniform 23x3.8 mm² cross section. Rod CF is made of aluminum (Ea=70 GPa) and has an uniform 31x3.5 mm² cross section. If a load P=19 kN were applied at A, determine the deflection at A. A 125 mm D B 225 mm 225 mm 150 mm E
Transcribed Image Text:Required information Deformable rods BD and CF are attached to a rigid rod ABC as shown. Rod BD is made of steel (Est-200 GPa) and has an uniform 23x3.8 mm² cross section. Rod CF is made of aluminum (Ea=70 GPa) and has an uniform 31x3.5 mm² cross section. If a load P=19 kN were applied at A, determine the deflection at A. A 125 mm D B 225 mm 225 mm 150 mm E
Equilib. Equations EMB=0→ 19x125-PCFX225-0-PCF-10.56(KN) (T) ΣFy=0→PBD=19+PCF-29.56(KN) (T) 4) Since PBD and PCF are (T), they will produce an increase in length of BD and CF respectively. Since end D is fixed, deflection of B is given as dB = SBD = PBD LBD EBD ABD PCP LCP ECP ACP 5) For rigid rod ABC, the magnitude of the deflection at A is given as 0.3275 Since end F is fixed, de flection of C is given as dc = $cF= 29.56(kN)x0.225 (m) 200x10 70x106 (mm) ( kN x23x3.8×10 (m²) 10.56(kN)x0.15(m) kN x31x3.5×10 (m²) () () mm = 0.38 ()(1) mm = 0.208
Transcribed Image Text:Equilib. Equations EMB=0→ 19x125-PCFX225-0-PCF-10.56(KN) (T) ΣFy=0→PBD=19+PCF-29.56(KN) (T) 4) Since PBD and PCF are (T), they will produce an increase in length of BD and CF respectively. Since end D is fixed, deflection of B is given as dB = SBD = PBD LBD EBD ABD PCP LCP ECP ACP 5) For rigid rod ABC, the magnitude of the deflection at A is given as 0.3275 Since end F is fixed, de flection of C is given as dc = $cF= 29.56(kN)x0.225 (m) 200x10 70x106 (mm) ( kN x23x3.8×10 (m²) 10.56(kN)x0.15(m) kN x31x3.5×10 (m²) () () mm = 0.38 ()(1) mm = 0.208
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