Eds - 5semestre
Por: laguiar.lima • 30/5/2016 • Trabalho acadêmico • 537 Palavras (3 Páginas) • 198 Visualizações
1
∑FH=0 → HÁ=0
∑FV=0 →VA-VB -8+3=0
VA-VB=5 → VA=5+0,3 VA = 5,5 Tf
∑MA=0 -3*6+4VB+2*8=0
-18+4VB+16=0
VB=2/4 → VB=0,5 Tf
Resp. C
2
∑FH=0 → -HÁ+20=0 → HÁ=20tf
∑FV=0 → VA-VB-8-3=0
VA-VB=5 → VA=5+0,5 → VA = 5,5tf
∑MA=0 → -3*6+4*VB+8*2=0
-18+4VB+16=0
VB=2/4 → VB=0,5 Tf
Resp. A
3
∑FH=0 → +10+15=HB → HB=25KN
∑FV=0 → -40-10+VA+VB=0
VA+VB=50
∑MB=0 → +2VA-10*2+5-40*1+10*2=0
+2VA-20+5-40+20=0
VA=35/2 → VA=17,5KN
+15*2+10*4+40*1-2HB-2VB+5=0
30+40+40-50-2VB+5=0
VB=32,5KN
Resp. C
4
∑FH=0 → HA=10KN
∑FV=0 → VA+VB-10-20+10=0
VA-VB=20
VB+10-10-20=0
VB=20
Resp.D
5
RESUMIDO
VB=140,9/2,73
VB=51,6KN
RA=RB-52,5/0,5
RA=51,6-52,5/0,5
RA=-18KN
Resp.E
HB-77,5-51,6
HB=25,9KN
6
MS2=4*2-5*2-10+4
MS2=12KN.M
∑FH=0 → HB=5KN
∑FV=0 → -4+R → R=4KN
∑MA=0 → 4+5*2-4*1=M
M=10KN.M
MS1=+2(1,2)²/4=1,44KN.M
MS3=-4*2+5*2+10-4 = 8
MS4= -4*2+10=2KN.M
Resp. A
7
RA=17,5 ↑
HB=15+10 →HB=25KN
RA+RB=50
∑MA=0 → 15*2+10*4-25*2-RB*2+40*1+5=0
RB=32,5KN
RA=50-32,5
RA=17,5KN
Resp.C
8
CONFERIDO A RESPOSTA BATENDO RESULTADO ESSA FOI A QUE MAIS SE APROXIMOU
Resp.E
9
HB=0
RA+RB=40+10+10
RA+RB=60 → RA=60-30 → RA=30KN
∑MA=0 → -10*2+40*1-2RB+10*4=0
RB=60/2
RB=30KN
Resp. B
10
∑FH=0 → RA*COS60°-15+5*SEN36,87°
RA=15-5*SEN36,87°/COS60°
RA=24KN
HA=24*COS60° HA=12KN
∑MA=0 → 20+20*2-RB*8+4*4+3*4=0
RB=88/8
RB=11KN
∑FV=0 → 11-4-20+VA=0
VA=13KN
Resp.A
11
∑FH=0 → HA = 2KN
M=2*10 → M=20KN.M
Resp.C
12
HB=15+10 HB=25KN
RA+RB=50
RA=50-32,5 → RA=17,5
∑MA=0 → 15*2+10*4+5+40*1-25*2-RB*2=0
RB=65/2 → RB=32,5KN
Resp.A
13
HA=20KN
RA=10KN
-10*4-20*2
-40-40
M= -80KN.M
Resp.C
14
H=0
RA+RB=120
∑MA=0 → -5,5RB+50+3,5+30*2+40*1=0
RB=50
RA=120-50 → RA=70KN
Resp.E
15
∑FH=0 → +HA+RB-5=0 → 4+RB-5=0
RB=5-4 → RB=1tf
∑MB=0 → -HA*3-2*3+3+5*3=0
-HA*3-6+3+15=0
HA=12/3
HA=4tf
Resp.C
16
M→P/X=3M
M25*3
M=75KN.M
Resp.A
17
HÁ=2KN
M=32KN.M
RA=5KN
Resp.A
18
∑FH=0 → HA=0
∑FV=0 → RA+RB=14 → RA=14-10 → RA=4KN
∑MA=0 → +4*5-4RB+10*2=0
20-4RB+20=0 → 4RB=20 →RB=40/4 → RB=10KN
Resp.A
19
∑FH=0 → -5+HA=0 → HA=5tf
∑FV=0 → RA-4=0 → RA=4tf
∑MA=0 → M=+4-5*2-4*1=0
M=-10tf.m
Resp.D
20
V=13-4X
13-4X=0
4X=13
X=13/4 X=3,25M
M=-8+13*3,25-4(3,25)²/2
M=13,1KN.M
P/X=4
M=-8+13*4-4(4)²/2
M=12KN.M
Resp.C
21
RA+RB=100
∑MA=0 → +6RB-100*3=0
RB=300/6 → RB=50KN
RA=50KN
M=10*5²/2-50*3=0
M=125-150
M=-25KN.M
Resp.B
22
TRECHO AB= A+B/2*H → 14+36/2*16 → AB=400 CM²
TRECHO BD=Πd²/4 → π(33)²/4 → BD=855,3 CM²
TRECHO DE=21*21 → DE = 441CM²
F/A= 1000/400 = 0,25KN.CM² → 250N/CM² → 2,5MPA TRAÇÃO
F/A = 100/855,3 = 0,12KN/CM² → 120N/CM² → 1,2MPA
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