**UFBA**

# Solucionáro Mecanica vetorial para engenheiros - Capitulo 7

PROBLEM 7.1

Determine the internal forces (axial force, shearing force, and bending moment) at point J of the structure indicated: Frame and loading of Prob. 6.7.

PROBLEM 7.2

Determine the internal forces (axial force, shearing force, and bending moment) at point J of the structure indicated: Frame and loading of Prob. 6.76.

PROBLEM 7.3

For the frame and loading of Prob. 6.80, determine the internal forces at a point J located halfway between points A and B.

SOLUTION FBD Frame:

0:80 kN0yyFAΣ=−= | 80 kNy=A |

PROBLEM 7.4

For the frame and loading of Prob. 6.101, determine the internal forces at a point J located halfway between points A and B.

SOLUTION FBD Frame:

0:100 N0yyFAΣ=−= | 100 Ny=A |

4.0 Nm=⋅M W

PROBLEM 7.5 Determine the internal forces at point J of the structure shown.

SOLUTION FBD Frame:

5 |

AB is two-force member, so yx yx A==

PROBLEM 7.6 Determine the internal forces at point K of the structure shown.

SOLUTION FBD Frame:

624 Nx=A AB is two-force member, so

A=→= | 260 Ny=A |

0:0xxxFACΣ=−+= | 624 Nxx==CA |

PROBLEM 7.7

A semicircular rod is loaded as shown. Determine the internal forces at point J.

SOLUTION FBD Rod:

PROBLEM 7.8

A semicircular rod is loaded as shown. Determine the internal forces at point K.

SOLUTION FBD Rod:

0:30 lb0yyFBΣ=−= | 30 lby=B |

0:20AxMrBΣ== | 0x=B |

PROBLEM 7.9

An archer aiming at a target is pulling with a 210-N force on the bowstring. Assuming that the shape of the bow can be approximated by a parabola, determine the internal forces at point J.

SOLUTION FBD Point A:

130:2210 N05xFTΣ=−= | 12175 NTT== |

By symmetry 12TT=

At :0.64 m, | 0.16 mBxy== ()2 |

Slope of parabola tan2dyaxdx θ===

PROBLEM 7.9 CONTINUED

50.4 Nm=⋅M W

PROBLEM 7.10

For the bow of Prob. 7.9, determine the magnitude and location of the maximum (a) axial force, (b) shearing force, (c) bending moment.

SOLUTION FBD Point A:

By symmetry | 12TTT== |

1130:2210 N0 | 175 N5xFTTΣ=−== |

()40:175 N0 | 140 N5yCCFFFΣ=−== |

()30:175 N0 | 105 N5xCCFVVΣ=−== |

And 11tantan2dyaxdx θ−−==

And | ()()105 Nsin140 NcosdVd |

θθθ=−+ |

Therefore, | (a) max140.0 N=F at C W |

(b) | max156.5 N=V at B W |

(c) max89.6 Nm=⋅M at C W

PROBLEM 7.1

A semicircular rod is loaded as shown. Determine the internal forces at point J knowing that o30.=θ

PROBLEM 7.12

A semicircular rod is loaded as shown. Determine the magnitude and location of the maximum bending moment in the rod.

for | 1tan882.87θ−==°, |

So 565 lbin. at 82.9 is a for MmaxACθ=⋅=°

()560 lbin.sin0 | for 0, where 0dMMdβββ=⋅=== |

PROBLEM 7.13

Two members, each consisting of straight and 168-m-radius quartercircle portions, are connected as shown and support a 480-N load at D. Determine the internal forces at point J.

SOLUTION FBD Frame:

()24240:0 | 375 N360 N |

PROBLEM 7.14

Two members, each consisting of straight and 168-m-radius quartercircle portions, are connected as shown and support a 480-N load at D. Determine the internal forces at point K.

96 Ny=C

0:0 | 96 NyyFBCΣ=−==B |

PROBLEM 7.15

Knowing that the radius of each pulley is 7.2 in. and neglecting friction, determine the internal forces at point J of the frame shown.

SOLUTION FBD Frame:

FBD BCE with pulleys and cord:

0:140 lb0 | 140 lbxxxFEΣ=−==E |

2.5 lby=E

PROBLEM 7.16

Knowing that the radius of each pulley is 7.2 in. and neglecting friction, determine the internal forces at point K of the frame shown.

SOLUTION FBD Whole:

2.5 lby=E

PROBLEM 7.17

Knowing that the radius of each pulley is 7.2 in. and neglecting friction, determine the internal forces at point J of the frame shown.

SOLUTION FBD Whole:

FBD BE with pulleys and cord:

FBD JE and pulley:

150 lby=B

30 lby=E

PROBLEM 7.18

Knowing that the radius of each pulley is 7.2 in. and neglecting friction, determine the internal forces at point K of the frame shown.

SOLUTION FBD Whole:

60 lbyA=− | 60 lby=A |

0:xFΣ= | 0=FW |

PROBLEM 7.19

A 140-m-diameter pipe is supported every 3 m by a small frame consisting of two members as shown. Knowing that the combined mass per unit length of the pipe and its contents is 28 kg/m and neglecting the effect of friction, determine the internal forces at point J.

SOLUTION FBD Whole:

FBD pipe: FBD BC:

By symmetry: 12NN= |

Also note: 20tan70 mm21 |

FBD CJ: PROBLEM 7.19 CONTINUED

PROBLEM 7.20

A 140-m-diameter pipe is supported every 3 m by a small frame consisting of two members as shown. Knowing that the combined mass per unit length of the pipe and its contents is 28 kg/m and neglecting the effect of friction, determine the internal forces at point K.

SOLUTION FBD Whole:

FBD pipe FBD AD:

FBD AK: PROBLEM 7.20 CONTINUED

PROBLEM 7.21

A force P is applied to a bent rod which is supported by a roller and a pin and bracket. For each of the three cases shown, determine the internal forces at point J.

SOLUTION (a) FBD Rod:

FBD AJ: (b) FBD Rod:

2 P=V W

2 aP=M W

FBD AJ: (c) FBD Rod:

2P=F W

PROBLEM 7.2

A force P is applied to a bent rod which is supported by a roller and a pin and bracket. For each of the three cases shown, determine the internal forces at point J.

SOLUTION (a) FBD Rod:

(b) FBD Rod:

2 P=F W

0:Σ=JM0=M W

(c) FBD Rod: FBD AJ:

PROBLEM 7.2 CONTINUED

PROBLEM 7.23

A rod of weight W and uniform cross section is bent into the circular arc of radius r shown. Determine the bending moment at point J when θ = 30°.

FBD CJ: |

PROBLEM 7.24

A rod of weight W and uniform cross section is bent into the circular arc of radius r shown. Determine the bending moment at point J when θ = 120°.

SOLUTION (a) FBD Rod:

0.1788Wr=M W

PROBLEM 7.25

A quarter-circular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at point J when o30.θ=

SOLUTION FBD Rod:

15,α=° 30weight of segment

PROBLEM 7.26

A quarter-circular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at point J when o30.θ=

SOLUTION FBD Rod:

Weight of segment 30

0.289Wr=M W

PROBLEM 7.27

For the rod of Prob.7.26, determine the magnitude and location of the maximum bending moment.

SOLUTION FBD Bar:

sin,rαα=

Weight of segment 2/2 Wαπ=

2 sin cos WF W

()24 sin cos sin cosrM Wr Wαθθ θ α απα π =+ −

But, 11sin cos sin2 sin22 αααθ==

so ()2sincossinWrMθθθθπ=+− |

or 2cosMWrθθπ=

()2 cos sin 0 at tan 1dM Wrd =−==θθθθθθπ

PROBLEM 7.27 CONTINUED

Solving numerically 0.8603radand0.357Wrθ==M W at 49.3θ=° W

(Since0M= at both limits, this is the maximum)

PROBLEM 7.28

For the rod of Prob.7.25, determine the magnitude and location of the maximum bending moment.

SOLUTION FBD Rod:

,s in2 θααα==r

Weight of segment 24/2 Wααππ==

()24 1 cos cos sin cosWW rMr αθ θθ α αππ α=+ − −

But, 11sincossin2sin22 αααθ==

so ()21coscossinθθθθπ=−+−rMW

()2 sin sin cos cos 0dM rWd θθθθθθπ=−+−=

At 0,Mθ== | at 1radθ= |

Only 0 and 1 in valid range at 57.3θ=° max0.1009MMWr==W

PROBLEM 7.29

For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

FBD beam:

22yLADw== | 4ywL==AD |

Σ= − = =y wLwLFVV

Σ= − = =J wLwLMMxMx

Along BC:

straight with | 10at4 |

xLwLMMwxx

PROBLEM 7.29 CONTINUED

LMwLx==

Section CD by symmetry (b) From diagrams:

maxon and 4 =wLVABCDW

2max3at center32 =wLMW

PROBLEM 7.30

For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION (a) Along AB:

straight with at22 =−=wLLVx

parabola with | 2 |

Along BC:

wLLMx

straight with | 213at82 |

=−=LMwLx

(b) From diagrams: | maxon 2 |

=wLVBCW

PROBLEM 7.31

For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION (a) Along AB:

0: 0 | yFPVVPΣ=−== |

0: 0 | JMMPxMPxΣ=−== |

straight with at 2 |

Along BC:

0: 0 | 0Σ=−−==yFPPVV |

(b) From diagrams: | maxalong VPAB=W |

maxalong 2 PLMBC=W

PROBLEM 7.32

0y ML =A

0:0Σ=−+=yyFAC | 0ML |

Along AB:

0:0 | Σ=−−==−y |

0: 0 | Σ=+==−JMMMxMMxLL |

straight with 0at 2 MMB=−

Along BC:

0: 0 | y |

0: 0 | 1K |

straight with 0at | 0 at 2 |

(b) From diagrams: maxVP= everywhere W

0max at 2 MMB=W

PROBLEM 7.3

SOLUTION (a) FBD Beam:

Along AC:

Along DE: Along EB:

PROBLEM 7.3 CONTINUED Along DE:

()0:12.625108 kips0 | 5.375 kipsyFVVΣ=−−−==− |

0:9.375 kips0 | 9.375 kipsyFVVΣ=+== |

(b) From diagrams: | max12.63 kips on VAC=W |

max29.8 kipft at MD=⋅W

PROBLEM 7.34

13.5 kNy=C Along AC:

()0:4 kN0 | 4 kN JMMxMxΣ=+==− |

4.8 kNm at MC=−⋅

Along CD:

PROBLEM 7.34 CONTINUED Along DE:

Along EB:

()21.5 kN | 1.8 kNm at MxME=−=−⋅ |

(b) From diagrams: | max9.50 kNVonCD=⋅W |

PROBLEM 7.35

SOLUTION (a) Along AC:

0:3 kip0 | 3 kipsyFVVΣ=−−==− |

()()0: 3 kips0 | 3 kipsJMMxMxΣ=+== |

9 kipft at MC=−⋅

Along CD:

0:3 kips5 kips 0 | 8 kipsyFVVΣ=−−−==− |

PROBLEM 7.35 CONTINUED Along EB:

0:3 kips5 kips 6 kips4 kips0 | 6 kipsyFVVΣ=−−+−−==− |

(b) From diagrams: | max8.0 kips on VCD=W |

max33.0 kipft at MB=⋅W

PROBLEM 7.36

540 N0 | 1350NyyE−+==E |

()()0: 540 N0 | 540 NJMxMMxΣ=+==− |

Along CD:

0:540 N1080 N0 | 540 NyFVVΣ=−+−== |

PROBLEM 7.36 CONTINUED Along DE:

0:1350 N540 N0 | 810 NyFVVΣ=+−==− |

0:540 N0 | 540 NyFVVΣ=−== |

()220:540 N0 | 540 N LMMxMxΣ=+==− |

()2162 Nm at | 0.3 mMEx=−⋅= |

(b) From diagrams | max810 N on VDE=W |

max162.0 Nm at and MDE=⋅W

PROBLEM 7.37

2 kipsy=A

114 kipftA=⋅M Along AC:

0:12 kips2 kips0 | 14 kipsyFVVΣ=−−== |