Let A and B these are the ends of the beam and \text{Let A and B these are the ends of the beam and } Let A and B these are the ends of the beam and
point C is the middle of the beam, then \text{point C is the middle of the beam, then} point C is the middle of the beam, then
A P = 1 ; P C = 2 ; P Q = 4 ; P B = 5 ; AP = 1;PC=2;PQ=4;PB = 5; A P = 1 ; PC = 2 ; PQ = 4 ; PB = 5 ;
By the condition of the problem: \text{By the condition of the problem:} By the condition of the problem:
F ⃗ a = − 100 ; F ⃗ c = − 40 ; F ⃗ b = − 80 ; \vec F_a= -100;\vec F_c=-40;\vec F_b=-80; F a = − 100 ; F c = − 40 ; F b = − 80 ;
let V p and V q support reactions \text{let } V_p \text{ and }V_q\text{ support reactions} let V p and V q support reactions
Σ M p = 0 ; Σ M q = 0 ; \Sigma M_p=0;\Sigma M_q=0; Σ M p = 0 ; Σ M q = 0 ;
Σ M p = − F a ∗ A P + F c ∗ P C + F b ∗ P B − V q ∗ P Q ; \Sigma M_p = - F_a * AP + F_c * PC + F_b* PB - V_q * PQ ; Σ M p = − F a ∗ A P + F c ∗ PC + F b ∗ PB − V q ∗ PQ ;
Σ M p = − 100 + 40 ∗ 2 + 80 ∗ 5 − 4 ∗ V q \Sigma M_p= -100+40*2+80*5-4*V_q Σ M p = − 100 + 40 ∗ 2 + 80 ∗ 5 − 4 ∗ V q ;
V q = 95 N V_q =95 N V q = 95 N
Σ M q = − F a ∗ A Q − F c ∗ P C + F B ∗ P B + V p ∗ P Q ; \Sigma M_q=-F_a*AQ-F_c*PC+F_B*PB+V_p*PQ; Σ M q = − F a ∗ A Q − F c ∗ PC + F B ∗ PB + V p ∗ PQ ;
V p = ( − 100 ∗ 5 − 40 ∗ 2 + 80 ∗ 1 ) − 4 = 125 N V_p=\frac{ ( - 100 * 5 - 40 * 2 + 80 * 1)} { - 4} = 125N V p = − 4 ( − 100 ∗ 5 − 40 ∗ 2 + 80 ∗ 1 ) = 125 N
Answer: V p = 125 N ; V q = 95 N V_p=125 N;V_q = 95N V p = 125 N ; V q = 95 N
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