Solución del problema 7.42

⤽ΣMA =0:10B − (15 kips)(3 ft) − (12 kips)(6 ft) = 0

B = +11.70 kips B =11.70 kips↑⤽ΣFy =0:Ay −15 −12 +11.70 =0

Ay = +15.30 kips A =15.30 kip↑0≤x≤6

⤽ΣM1=0: 1.25x2 -15.3x +M= 0

M =15.30x −1.25x2

↑ΣFy =0: 15.3-2.5x -V=0

V= 15.3-2.5x

6≤x≤10

⤽ΣM2=0: 15(x-3) -15.3x +M +12(x-6) = 0

M =15.30x −15(x-3) -12(x-6)↑ΣFy =0: 15.3-15-12 -V=0

V= -11.7

|V|max =15.30 kips

0 6V 15.30 0.30M 0 46.8

6 10V -11.7 -11.7M 46.8 0

|M|max = 46.8 kip ⋅ ft

Solución del problema 7.50

⤽ΣMA =0:-400(0.30+0.60+0.90)+0.9B=0

B=+800 B=800N↑↑ΣFy =0: -400x3+A+800 =0

A=+400 A=400 N↑0≤x≤0.15

⤽ΣM1 =0: -400X +M =0

M=400X

⤽ΣFy =0: 400-V =0

V=400N

0.15≤x≤0.45

⤽ΣM2 =0: -400X 400(X -0.15) -60+M =0

M=120N

⤽ΣFy =0: 400 -400-V =0

V=0

0.45≤x≤0.75

⤽ΣM3 =0: -400X 400(X -0.15) -120+ 400(X-0.45)+M =0

M=180 -400(X-0.45)

⤽ΣFy =0: 400 -400-400 –V=0

V=-400

0≤y≤0.15

0 0.15V 400 400M 0 60

0.15 0.45V 0 0M 120 120

0.45 0.75V -400 -400M 180 60

⤽ΣM4 =0: 800X+V=0

V=-800

⤽ΣFy =0: 800+V=0

V=-800

|V|max =800N

|M|max =180Nm

0 0.15V -800 -800M 120 0

Suloucion del problema 7.87

Solución del problema 7.89

En AD⤽ΣMD =0: -C(0.3 m) +0.800 kN m + (6 kN)(0.45 m) = 0

C =+11.667 kN C =11.667 kN↑En EB ⤽ΣMA =0: B(0.3 m) -1.300 kN m = 0

B = 4.333 kN

En D

⤽ΣMA =0: (6 kN) (0.45 m) -(11.667 kN)(0.3 m) +(0.3 m) (4.333 kN)(0.6 m) –Q.03m =0

Q= 6.00KN↓ ↑ΣFy =0: 11.667 kN +4.333 kN −6 kN − P − 6 kN = 0

P=4.00KN↓0≤x≤0.3

⤽ΣM1 =0: 10X2 +M=0

M=-10X2

↑ΣFy =0:-20X-V=0

V= 20X

0.30≤x≤0.60

⤽ΣM2 =0: 6(X-0.15)-11.667(X-0.30)+M=0

M=11.667(X-0.30)- 6(X-0.15)↑ΣFy =0:-6+11.667-V=

V= 5.667

0.30≤Y≤0.60

⤽ΣM3 =0: 4.333X-6(X-0.30)-M=0

0 0.30V 0 -6M 0 -900

0.03 0.60V 5.667 5.667M -900 800

M=4.333X-6(X-0.30) ↑ΣFy =0: 4.333-6+ V=0

V= 1.667

0≤Y≤0.30

⤽ΣM4 =0: 4.333X –M =0

M=4.333X↑ΣFy =0: 4.333 +V=0

V= -4.333

|V|max = 6KN

|M|max= 1300Nm

0.03 0.60V 1.667 1.667M 1300 800

0 0.30V -4.333 -4.333M 0 1300