Review Materials engineering

Problem 604604-resultant-3d-forces.gif
Determine the magnitude of the resultant, its pointing, and its direction cosines for the following system of non-coplanar concurrent forces. 200 lb (4, 5, 3); 400 lb (6, 4, 5); 300 lb, (4, 2, 3).

[MODE] 8:VECTOR 1:VctA 1:3
VctA = [ 4 5 -3 ]

[AC] [SHIFT] [5:VECTOR] 2:Data 2:VctB 1:3
VctB = [ -6 4 -5 ]

[AC] [SHIFT] [5:VECTOR] 2:Data 3:VctC 1:3
VctC = [ 4 -2 -3 ]

Resultant R
R = 200(VctA Abs(VctA)) + 400(VctB Abs(VctB)) + 300(VctC Abs(VctC))

[AC] 200 [ ( ] [SHIFT] [5:VECTOR] 3:VctA [ &divide] [SHIFT] [hyp:Abs] [SHIFT] [5:VECTOR] 3:VctA [ ) ] [ ) ] [ + ] 400 [ ( ] [SHIFT] [5:VECTOR] 4:VctB [ &divide] [SHIFT] [hyp:Abs] [SHIFT] [5:VECTOR] 4:VctB [ ) ] [ ) ] [ + ] 300 [ ( ] [SHIFT] [5:VECTOR] 5:VctC [ &divide] [SHIFT] [hyp:Abs] [SHIFT] [5:VECTOR] 5:VctC [ ) ] [ ) ]

R = [ 62.46 212.32 -479.91 ] answer

For the magnitude of the resultant:

[AC] [SHIFT] [hyp:Abs] [SHIFT] [5:VECTOR] 6:VctAns [ ) ].
R = Abs(VctAns) = 528.48 lb answer

For the direction cosines:

[AC] [SHIFT] [5:VECTOR] 6:VctAns [ ) ] [SHIFT] [hyp:Abs] [SHIFT] [5:VECTOR] 6:VctAns [ ) ].
= VctAns Abs(VctAns)

= [ 0.1182 0.4018 -0.9081 ] answer

- See more at: http://www.mathalino.com/reviewer/engineering-mechanics/problem-603-resultant-concurrent-forces-space4#sthash.I3VhbAVo.dpuf

Problem 603
Determine the magnitude of the resultant, its pointing, and its direction cosines for the following system of non-coplanar concurrent forces. 100 lb (2, 3, 4); 300 lb (3, 4, 5); 200 lb, (0, 0, 4). - See more at: http://www.mathalino.com/reviewer/engineering-mechanics/problem-603-resultant-concurrent-forces-space#sthash.XMaGcmZk.dpuf

[MODE] 8:VECTOR 1:VctA 1:3
VctA = [ 2 3 4 ]

VctB = [ -3 -4 5 ]

VctC = [ 0 0 4 ]

Resultant R


R = [ -90.14 -114 486.41 ] answer


R = Abs(VctAns) = 507.66 lb answer


= VctAns Abs(VctAns)

= [ -0.1776 -0.2246 0.9581 ] answer

- See more at: http://www.mathalino.com/reviewer/engineering-mechanics/problem-603-resultant-concurrent-forces-space#sthash.XMaGcmZk.dpuf

Centroids of Composite Figures

Center of gravity of a homogeneous flat plate

Wx=wx

Wy=wy

Centroids of areas

Ax=ax

Ay=ay

Centroids of lines

Lx=lx

Ly=ly

Center of Gravity of Bodies and Centroids of Volumes

Center of gravity of bodies

Wx=wx

Wy=wy

Wz=wz

Centroids of volumes

Vx=vx

Vy=vy

Vz=vz

Centroids Determined by Integration

Centroid of area

Ax=baxcdA

Ay=baycdA

Centroid of lines

Lx=baxcdL

Ly=baycdL

Center of gravity of bodies

Wx=baxcdW

Wy=baycdW

Wz=bazcdW

Centroids of volumes

Vx=baxcdV

Vy=baycdV

Vz=bazcdV

Centroids of Common Geometric Shapes

Rectangle

Area and Centroid

A=bd

x=12b

y=12d

Triangle

Area and Centroid

A=12bh

y=13h

Circle

Area and Centroid

A=r2

x=0

y=0

Semicircle

Area and Centroid

A=12r2

x=0

y=4r3

Semicircular Arc

Length and Centroid

L=12r2

x=2r

y=0

Quarter Circle

Area and Centroid

A=14r2

x=4r3

y=4r3

Sector of a Circle

Area and Centroid

A=r2rad

x=2rsin3rad

y=0

Circular Arc

Length and Centroid

L=2rrad

x=rsinrad

y=0

Ellipse

Area and Centroid

A=ab

x=0

y=0

Half Ellipse

Area and Centroid

A=12ab

x=0

y=4b3

Quarter Ellipse

Area and Centroid

A=14ab

x=4a3

y=4b3

Parabolic Segment

Area and Centroid

A=23bh

x=38b

y=25h

Spandrel

Area and Centroid

A=1n+1bh

x=1n+2b

y=n+14n+2h

- See more at: http://www.mathalino.com/reviewer/engineering-mechanics/centroids-and-centers-gravity#sthash.WuYiEh6P.dpuf

Problem 602
Determine the magnitude of the resultant, its pointing and its direction cosines for the following system of non-coplanar, concurrent forces. 300 lb (+3, -4, +6); 400 lb (-2, +4, -5); 200 lb (-4, +5, -3). - See more at: http://www.mathalino.com/reviewer/engineering-mechanics/problem-602-resultant-concurrent-forces#sthash.u6ZTQXMB.dpuf

MODE] 8:VECTOR 1:VctA 1:3
VctA = [ 3= -4= 6= ]

VctB = [ -2= 4= -5= ]

VctC = [ -4= 5= -3= ]

Resultant R
The calculator will display the right hand of the equation below. Press [ = ] to get the resultant.


R = [ -117.16 226.29 -152.53 ] answer


R = Abs(VctAns) = 296.98 lb


= VctAns Abs(VctAns) = [ -0.394 0.762 -0.514 ] answer - See more at: http://www.mathalino.com/reviewer/engineering-mechanics/problem-602-resultant-concurrent-forces#sthash.u6ZTQXMB.dpuf

Compound Interest

SPONSORED LINKS

Log in or Register to Discuss This!

In compound interest, the interest earned by the principal at the end of each interest period (compounding period) is added to the principal. The sum (principal + interest) will earn another interest in the next compounding period.

Consider $1000 invested in an account of 10% per year for 3 years. The figures below shows the contrast between simple interest and compound interest.

At 10% simple interest, the $1000 investment amounted to $1300 after 3 years. Only the principal earns interest which is $100 per year.

000-simple-interest.gif

At 10% compounded yearly, the $1000 initial investment amounted to $1331 after 3 years. The interest also earns an interest.

000-compound-interest.gif

Elements of Compound Interest
P = principal, present amount
F = future amount, compound amount
i = interest rate per compounding period
r = nominal annual interest rate
n = total number of compounding in t years
t = number of years
m = number of compounding per year

i=rm and n=mt

Future amount,

F=P(1+i)n or F=P(1+rm)mt

The factor (1+i)n is called single-payment compound-amount factor and is denoted by (F/P,i,n).

Present amount,

P=F(1+i)n

The factor 1(1+i)n is called single-payment present-worth factor and is denoted by (P/F,i,n).

Number of compounding periods,

n=ln(F/P)ln(1+i)

Interest rate per compounding period,

i=FPn1

Values of i and n
In most problems, the number of years t and the number of compounding periods per year m are given. The example below shows the value of i and n.

Example
Number of years, t=5years
Nominal rate, r=18%

Compounded annually (m=1)

n=1(5)=5

i=0.18/1=0.18

Compounded semi-annually (m=2)

n=2(5)=10

i=0.18/2=0.09

Compounded quarterly (m=4)

n=4(5)=20

i=0.18/4=0.045

Compounded semi-quarterly (m=8)

n=8(5)=40

i=0.18/4=0.0225

Compounded monthly (m=12)

n=12(5)=60

i=0.18/12=0.015

Compounded bi-monthly (m=6)

n=6(5)=30

i=0.18/6=0.03

Compounded daily (m=360)

n=360(5)=1800

i=0.18/360=0.0005

Continuous Compounding (m )
In continuous compounding, the number of interest periods per year approaches infinity. From the equation
F=(1+rm)mt

when m, mt=, and rm0. Hence,
F=Plimm(1+rm)mt

Let x=rm. When m, x0, and m=rx.

F=Plimx0(1+x)rxt

F=Plimx0(1+x)1xrt

From Calculus, limx(1+x)1/x=e, thus,

F=Pert

- See more at: http://www.mathalino.com/reviewer/engineering-economy/compound-interest#sthash.UlnX1aOi.dpuf

Problem 001-mm
The structure shown in Fig F-001(MM) is pinned together at points A, B, and C and held in equilibrium by the cable CD. A load of 12,000 lb is acting at the midpoint of member AB, and a load of 8000 lb is applied at point C. Determine the reaction at A, the internal force in member BC, and the tension on cable CD.

a=16cos30=13.86m

b=16sin30=8m

c=atan37=13.86tan37=10.44m

001-mm-free-body-diagram.gif

Tension on cable CD
MA=0

(Tsin53)(8+10.44)=8000(13.86)+12000(4)

T=10788.47lb answer

Reaction at A
MD=0

Ax(8+10.44)=8000(13.86)+12000(4)

Ax=8616.05lb

FV=0

Ay+Tcos53=8000+12000

Ay+10788.47cos53=8000+12000

Ay=13507.34lb

RA=Ay2+Ax2

RA=8616.052+13507.342

RA=16021.38lb

tanAx=AyAx=\defrac13507.348616.05

Ax=57.47

Thus, RA = 16 021.38 lb at Ax = 57.47 with the horizontal. answer

Force on member CD
tan=813.868

=53.78

001-mm-member-bc.gif

MA=0

(FBCsin)(8)=12000(4)

(FBCsin53.78)(8)=12000(4)

FBC=7437.21lb tension answer

- See more at: http://www.mathalino.com/reviewer/engineering-mechanics/problem-001-mm-method-members#sthash.vfS7n5ek.dpuf

Problem 007-cb
In the structure shown in Fig. CB-007(FR), members BCE, and CD are assumed to be solid rigid members. Members AE and DE are cables. For this structure, determine the
reaction at B.

Structure supported by cable

- See more at: http://www.mathalino.com/reviewer/engineering-mechanics/problem-007-cb-analysis-cabled-frame#sthash.hVzp7GWU.dpuf

MB=0

24(213T)=12000(10sin53)

T=7198.80lb

FH=0

BH=213T

BH=213(7198.81)

BH=3993.18lb answer

FV=0

BV=313T+12000

BV=17989.77lb answe

- See more at: http://www.mathalino.com/reviewer/engineering-mechanics/problem-007-cb-analysis-cabled-frame#sthash.hVzp7GWU.dpuf

Problem 006-fr
In the structure shown in Fig. P-006(FR-H), all members are assumed to be solid rigid members. The system is pinned to the wall at point A and supported by a roller at point E. Calculate the force on member BD and the reactions at A and E.

Simple Frame (Inverted L-frame)

Solution 006-fr

Hide Click here to show or hide the solution

MA=0

4RE=6(120)

RE=180kN answer

006-fr-free-body-diagram.gif

FH=0

AH=RE

AH=180kN answer

FV=0

AV=120kN answer

006-fr-fbd-member-abc.gif

MA=0

3(213FBD)=6(120)

FBD=432.67kN answer

- See more at: http://www.mathalino.com/reviewer/engineering-mechanics/problem-006-fr-analysis-simple-frame#sthash.KXLnKHeH.dpuf

Problem 104
A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. Determine the outside diameter of the tube if the stress is limited to 120 MN/m2.

Solution 104


104-hollow-tube.gifP=A

where:
P=400kN=400000N

=120MPa

A=14D214(1002)

A=14(D210000)

Thus,
400000=120[14(D210000)]

400000=30D2300000

D2=400000+30000030

D=119.35mm answer

- See more at: http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-104-normal-stress#sthash.vXYfLEqA.dpuf

]

Problem 105
A homogeneous 800 kg bar AB is supported at either end by a cable as shown in Fig. P-105. Calculate the smallest area of each cable if the stress is not to exceed 90 MPa in bronze and 120 MPa in steel.

105-bar-suspended-by-rods.gif

Solution 105


By symmetry:
Pbr=Pst=12(7848)

Pbr=3924N

Pst=3924N

105-fbd-of-bar.gif

For bronze cable:
Pbr=brAbr

3924=90Abr

Abr=43.6mm2 answer

For steel cable:
Pst=stAst

3924=120Ast

Ast=32.7mm2 answer

- See more at: http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-105-normal-stress#sthash.lUAGecCn.dpuf

Problem 108
An aluminum rod is rigidly attached between a steel rod and a bronze rod as shown in Fig. P-108. Axial loads are applied at the positions indicated. Find the maximum value of P that will not exceed a stress in steel of 140 MPa, in aluminum of 90 MPa, or in bronze of 100 MPa.

108-composite-bar-different-area.gif

Solution 108


For bronze:
brAbr=2P

100(200)=2P

P=10000N

108-stresses-composite-bar-different-area.gif

For aluminum:
alAal=P

90(400)=P

P=36000N

For Steel:
stAst=5P

P=14000N

For safe value of P, use the smallest above. Thus,
P=10000N=10kN answer

- See more at: http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-108-normal-stress#sthash.nXU2Qmls.dpuf

Problem 113
Find the stresses in members BC, BD, and CF for the truss shown in Fig. P-113. Indicate the tension or compression. The cross sectional area of each member is 1600 mm2.

113-cantilever-like-truss.gif

Solution 113


For member BD: (See FBD 01)
MC=0

3(45BD)=3(60)

BD=75kN Tension

113-fbd-01-section-cf-cd-bd.gif

BD=BDA

75(1000)=BD(1600)

BD=46.875MPa (Tension) answer

For member CF: (See FBD 01)
MD=0

4(12CF)=4(90)+7(60)

CF=275.77kN Compression

CF=CFA

275.77(1000)=CF(1600)

CF=172.357MPa (Compression) answer

For member BC: (See FBD 02)

113-fbd-02-section-ac-bc-bd.gif

MD=0

4BC=7(60)

BC=105kN Compression

BC=BCA

105(1000)=BC(1600)

BC=65.625MPa (Compression) answer

- See more at: http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-113-normal-stress#sthash.Ed2Rpk0n.dpuf

Forces parallel to the area resisting the force cause shearing stress. It differs to tensile and compressive stresses, which are caused by forces perpendicular to the area on which they act. Shearing stress is also known as tangential stress.

=VA

where V is the resultant shearing force which passes through the centroid of the area A being sheared.

000-single-shear-double-shear.gif

- See more at: http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/shear-stress#sthash.jZIqfssK.dpuf

Problem 115
What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm thick? The shear strength is 350 MN/m2.

Solution 115


The resisting area is the shaded area along the perimeter and the shear force V is equal to the punching force P.

115-punching-a-hole.gif

V=A

P=350[(20)(25)]

P=549778.7N

P=549.8kN answer

- See more at: http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-115-shear-stress#sthash.wlA1xSTU.dpuf

Problem 117
Find the smallest diameter bolt that can be used in the clevis shown in Fig. 1-11b if P = 400 kN. The shearing strength of the bolt is 300 MPa.

117-clevis.gif

Solution 117


The bolt is subject to double shear.
V=A

400(1000)=300[2(14d2)]

d=29.13mm answer

- See more at: http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-117-shear-stress#sthash.Ba3okDci.dpuf

Problem 120
The members of the structure in Fig. P-120 weigh 200 lb/ft. Determine the smallest diameter pin that can be used at A if the shearing stress is limited to 5000 psi. Assume single shear.

120-structure-in-three-hinges.gif

Solution 120


For member AB:

120-fbd-member-ab.gif

Length, LAB=42+42=5.66ft

Weight, WAB=5.66(200)=1132lb

MA=0

4RBH+4RBV=2WAB

4RBH+4RBV=2(1132)

RBH+RBV=566 Equation (1)

For member BC:

120-fbd-member-bc.gif

Length, LBC=32+62=6.71ft

Weight, WBC=6.71(200)=WBC=1342lb

MC=0

6RBH=1.5WBC+3RBV

6RBH3RBV=1.5(1342)

2RBHRBV=671 Equation (2)

Add equations (1) and (2)
RBH+RBV=566 Equation (1)

2RBHRBV=671 Equation (2)

3RBH=1237

RBH=412.33lb

From equation (1):
412.33+RBV=566

RBV=153.67lb

From the FBD of member AB
FH=0

RAH=RBH=412.33lb

FV=0

RAV+RBV=WAB

RAV+153.67=1132

RAV=978.33lb

RA=R2AH+R2AV

RA=412.332+978.332

RA=1061.67lb shear force of pin at A

V=A

1061.67=5000(14d2)

d=0.520in answer

- See more at: http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-120-shear-stress#sthash.Aljlu90g.dpuf

Problem 122
Two blocks of wood, width w and thickness t, are glued together along the joint inclined at the angle as shown in Fig. P-122. Using the free-body diagram concept in Fig. 1-4a, show that the shearing stress on the glued joint is = P sin 2 / 2A, where A is the cross-sectional area.

122-block-of-wood.gif

122-forces-on-arbitrary-section.gif

Solution 122


Shear area,
Ashear=t(wcsc)

Ashear=twcsc

Ashear=Acsc

Shear force,
V=Pcos

122-fbd-normal-parallel-components.gif

V=Ashear

Pcos=(Acsc)

=PsincosA

=P(2sincos)2A

=Psin22A (okay!)

- See more at: http://www.mathalino.com/reviewer/mechanics-and-strength-of-materials/solution-to-problem-122-shear-stress#sthash.wOCfumaG.dpuf

Review Materials engineering

Documents

Transcript of Review Materials engineering