MOS-I (Lecture 1-2) Engr. Khalid Yousaf

7/23/2019 MOS-I (Lecture 1-2) Engr. Khalid Yousaf

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MECHANICS OF SOLIDS-I

Simple Stress and Strain: Kinds of stresses and strains, Difference between stress andExtension Diagrams for different Materials, Hooks Law, Moduli of elasticity, Lateral sStrain, Poissons Ratio, Temperature stresses and Compound bars.

Analysis of Beams: Shear force and bending moment diagrams of beams under differe

conditions, Theory of simple bending, Moment of resistance and section modulus, Appformula, Shear Stresses in Beams, Shear Centre, Shear Flow.

Column and Struts: A short and long axially loaded columns, their modes of failure, equivalent length, Eulers formula, and Empirical formula like Rankine Gordon FormuRatio.

Circular Shafts: Theory of Torsion for solid and hollow circular shafts.

Springs: Open coil springs, closed coil springs, leaf springs.Strain Energy: Strain Energy due to direct loads, shear force, bending moments, torqu

Recommended Books:

1. Pytel, A. & F. L.Singer, Strength of Material, Harper & Row Publishers, New York.

2. Hibbler, R. C., Mechanics of Materials, Prentice Hall, 6th Edition, 2004.

3. Warnock, F. V., Benham, P. P., Mechanics of Solids and Strength of Materials, Pitman Publishing, 1970.

4. James M. Gere & Barry. J. Goodno, Mechanics of Materials, 7th Edition, 2008, CL Engineering

5. James M. Gere & Stephen P. Timoshenko, Mechanics of Materials, 4th Edition, 1997, PWS Pub Co.


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Normal Stresses: Stress is defined as the strength of a material per unit area or unit stremember divided by area, which carries the force, formerly express in psi, now in N/mm2 or MPa. Mathem

=

where P is the applied normal load in Newton and A is the area in mm 2. The maximum stress in tensio

over a section normal to the load.

Normal stress is either tensile stress or compressive stress.Members subject to pure tension (or tensile force) is under

tensile stress, while compression members (members

subject to compressive force) are under compressive stress.

Compressive force will tend to shorten the member.

Tension force on the other hand will tend to lengthen the member.


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Problem#101:The bar ABCD in Fig. (a) consists of three cylindrical steel segments with different lenareas. Axial loads are applied as shown. Calculate the normal stress in each segment.

Solution:We begin by using equilibrium analysis to

compute the axial force in each segment of the bar.

Step#1:Considering section 1 and summing up the forces

acting on it.= 4000 lb

=1.2 in2

=

=

=

=

.= 3333.3 ()

Step#2:Adding up forces in section 2, we get

= - 5000 lb

=1.8 in2

=

=

.

= 2777.8 =2777.7 Psi (C)


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Step#3:Taking section 3 under consideration.

= 7000 lb

=1.6 in2

=

=

.

= 4375 ()


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1

Problem 102:For the truss shown in theshown in the figure. Determine the stress in

the member AC and BD. The cross sectional

area of the each member of the truss is 900 mm2.

Solution:The three assumptions taken in the

elementary analysis of the truss are ,1- Weights of the members are neglected.2- All connections are smooth pins.3- All external loadsare applied directly to the pins.Step#1:Calculate the reactions at each support Fy =0Ay+ Hy= 50 kN

M

A=0

30 4 70 12 Hy 16 = 0Hy= 60 kNAndAy= 40 kNStep#2:Considering Section 1Fy =0

Ay+

FAB= 0

FAB=

AyFAB=

40 = 66.7 kN


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Fy =0

FAC+

FAB= 0

FAC=

FAB

FAC=

(66.7) C

FAC=53.4

Step#3:

ME =0

Ay 8 + FBD 3304 = 0

40 8 + FBD 3304 = 0

3FBD=200

FBD=66.7 kN

Step#4:

The area of each member = 900 mm2 =900 106 m2

The stresses in members AC and BD are, =

=53.4103

900106

= 59.3 106

= 59.3MPa (T)

=66.7103

900106

= 74.1 106

= 74.1MPa =74.1MPa (C)


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Problem 103:Figure (a) shows a two-member truss supporting a block of weight W. The cross-sectional areas of th

AB and 400 mm for AC . Determine the maximum safe value of W if the working stresses are 110 MPa for AB and 12

Solution.Being members of a truss, AB and AC can

be considered to be axially loaded bars . The forces

in the bars can be obtained by analyzing the FBD of

pin A in Fig. (b). The equilibrium equations are,

Fx =0

PAC 60- PAB 40

= 0

PAC 60+ PAB 40

- W= 0

Solving above equations simultaneously,

PAC = 0.7779PAB = 0.5077

Area of each member A = 800 mm2

= 800 106

m2

Stress in each member:=

=

120 106 =0.7779W800 106

W = 123409 N = 123 kN=

110 106 =0.5077W

800 106

W = 173330 N = 173 kN For safe loading, use W= 123 kN


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Alternatively:Make a polygon with the force vectors as shown in the figure.

Applying Law of Sine to this polygon , we have

For Member AC:

sin 50= W

sin 100

=0.7779W

=0.7779W120(800 106)=0.7779W

W = 123409 N =123 kN

For wire AB:

sin 30= W

sin 100

=0.5077W

=0.5077W

110 106(800 106)=0.5077W

W = 173330 N = 173 kNFor safe load W use W = 123 kN


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Problem 104A hollow steel tube with an inside diameter of 100 mm must carry a tensile load of 400 kN. De

diameter of the tube if the stress is limited to 120 MN/m2.Solution:Given that

Inside diameter d = 100 mm

Tensile load P = 400 kN = 400000 NStress = 120 MN/m2

Outer diameter D = ?

Since

=

or = A

Area = (D2 d2)

400000 = 120(D2 d2)

400000 = 30 (D2 1002)

=+

= 119.35


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Problem 105:A homogeneous 800 kg bar AB is supported at either end by a cable as shown in Fig. P-105. Caeach cable if the stress is not to exceed 90 MPa in bronze and 120 MPa in steel.Solutions:Given thatMass of the bar M = 800 kg

W = 7848 N

Stress in bronze br = 90 MpaStress in steel st = 90 Mpa

By symmetry:Pbr =Pst =12(7848)Pbr=3924 NPst=3924 NFor bronze cable:

Pbr= brAbr3924 = 90AbrAbr= 43.6 mm2

For steel cable:

Pst = stAst3924 = 120Ast

Ast= 32.7 mm2


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Problem106:The homogeneous bar shown in Fig. P-106 is supported by a smooth pin at Cfrom A to B around the smooth peg at D. Find the stress in the cable if its diameter is 0.6 in

6000 lb.

Solution:Given that

Diameter of the cable d = 0.6 in

Weight of the bar w = 600 lbMC = 0

5T+10(

T) = 5(6000)

T=2957.13 lbT = A

2957.13 = [

(0.62)]

= 10458.72 psi


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Problem 107:A rod is composed of an aluminum section rigidly attached between steel ashown in Fig. P-107. Axial loads are applied at the positions indicated. If P = 3000 lb and the

the rod is 0.5in2, determine the stress in each section.

Solution:Given that

Applied load P = 3000 lb = 3 kips

cross-Sectional Area A = 0.5in2

For steel:

stAst = Pstst(0.5) = 4P

st(0.5) =4 3

st = 24 ksi

For aluminum:

alAal = Palal(0.5) = 4P

al(0.5) = 4 3

al = 24 ksi

For bronze:

brAbr = Pbr

br(0.5) = 9

br = 18 ksi


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Problem 108:An aluminum rod is rigidly attached between a steel rod and a bronze rod asAxial loads are applied at the positions indicated. Find the maximum value of P that will not ex140 MPa, in aluminum of 90 MPa, or in bronze of 100 MPa.Solutions:Given thatStress in steel = 140 MpaSteel area = 500 mm

2

Stress in aluminum = 90 MpaAluminum area = 400 mm

2

Stress in bronze = 100 MpaBronze area = 200 mm

2

Max. applicable load P = ?For bronze:brAbr = Pbr1000(200) = 2PP = 10000 NFor aluminum:alAal = Pal90(400) = PP = 36000 NFor steel:stAst = Pst

100(500) = 5PP = 10000 N For safe value of P , use P = 10000 N = 10 kN


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Problem 109:Determine the largest weight W that can be supported by two wires shown in Feither wire is not to exceed 30 ksi. The cross-sectional areas of wires AB and AC are 0.4 in2 andSolution:Given thatMaximum applicable stress = 30 ksi

Cross-sectional area of AB = 0.4 in2

Cross-sectional area of AB = 0.5 in2

Largest supportable weight W = ?The free body diagram of the joint Ais given byFor wire AB:By sine law (from the force polygon):

sin 40= W

sin 80=0.6527W=0.6527W

30(0.4)=0.6527WW = 18.4 kipsFor wire AC:

sin 60= W

sin 80=0.8794W=0.8794W30(0.5)=0.6527W

W = 17.1 kips For safe load W use W = 17.1 kips


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Problem 110:A 12-inches square steel bearing plate lies between an 8-inches diameter wood

footing as shown in Fig. P-110. Determine the maximum value of the load P if the stress in w

psi and that in concrete to 650 psi.

Solution:Given that

Area of steel bearing plate = 12 in2

Diameter of wooden post = 8 inStress in wood = 1800 Psi

Stress in concrete = 650 Psi

For Wood:

Pw = wAwPw = w

d2

Pw = 1800[82]

Pw =90477.9 lbFor concrete:

PC= CAC

PC= 650 122

PC= 93600 lb.

For safe load P use P = Pw = 90477.9 lb


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Problem 111:For the truss shown in Fig. P-111, calculate the stresses in members CE, D

sectional area of each member is 1.8 in2. Indicate tension (T) or compression (C).

Solution:Given that

Cross Sectional Area of each member A = 1.8 in2

CE= ?

DE= ?

DF= ?= 0

RA+ RF=30 (1)

= 0

24 RF =16(30)

RF=20k put in (1)RA= 10k

At joint F:

= 0

-

DF= 20

DF = -33.33k = 33.33k (C)

Atjoint D:

By symmetry


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BD = DF = 33.33k (C)

= 0

=

DF

=

33.33

33.33

=40k (T)

At joint E:

= 0

30 = 40

CE = 10.67k (T)

Stresses:(Stress = Force/ Area)

E= .

.

= 5.93 (T)=

.

= 22.22 ()

= .

.

= 18.52 ()


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Problem#112:Determine the cross-sectional areas of members AG, BC, and CE for the trus

The stresses are not to exceed 20 ksi in tension and 14 ksi in compression. A reduced str

specified to reduce the danger of buckling.

Soln.

= 0

= 40 25 = 65 = 0

1 8 = 40 4 25 8

= 20

= 0

= = 20

Check:

= 01 2 = 18 40 8 25 4

12 65 = 18 20 40 8 25 4

780 . = 780 . (ok!)

For memberAG (At joint A):

= 0

= 65

AB = 78.12 k


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= 0

20 =

20 =

78.12

= 20.33 ()since

=

=

20= .

= 1.17

For member BC (At section through MN):= 0

1 2 = 6

12 20 = 6

B = 72.11 = 72.11 ()

=

14 = .

= 5.15


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For member CE (At joint D):= 0

20 =

CD = 36.06 k

= 0

=

=

36.06

= 30

At joint E:= 0

30 =

= 36.06

= 0

C =

C =

36.06

= 20 = 20 ()

=

14 =

= 1.43


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Problem#113:Find the stresses in members BC, BD, and CF for the truss shown in Fig. P-113

compression. The cross sectional area of each member is 1600 mm2.

Soln.

For member BD: (See FBD 01)

= 0

3

= 3 60 = 75 (T)

=

= ()

= 46.875 Mpa (T)

For member CF: (See FBD 01)

= 04

= 4 900 7 60

= 275.77 ()

=

= .()

= 172.375 ()


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For member BC: (See FBD 02)

= 0

4 = 7 60

= 105 ()

=

= ()= 62.625 ()


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Problem#114:The homogeneous bar ABCD shown in Fig. P-114 is supported by a cable that rthe smooth peg at E, a vertical cable at C, and a smooth inclined surface at D. Determine the mthat can be supported if the stress in each cable is limited to 100 MPa. The area of the cable Aof the cable at C is 300 mm2.Soln.= 0

300

= Sin300

= 1.1305= 0 6 Sin30

0 Cos500 =

Sin300 1.1305Cos50

0 = 2.226= = 2.226= 0

4 2 6 Sin300 = 37 2 ( 2.226) = 32.5466= = 0.3927= 2.226= 2.2267(0.3927)= 0.1256


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Based on cable AB:

=

100 =.

= 63661.83

Based on cable at C:

=

100 =.

= 238853.5

Safe value of W

= 63661.83

=

63661.83 = (9.81)

= 6490

= 6.49


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Assignment.

Q#1:Axial loads are applied to the compound rod that iscomposed of an aluminum segment rigidly connected

between steel and bronze segments. What is the stress ineach material given that P = 10 kN?

Q#2:Axial loads are applied to the compound rod that iscomposed of an aluminum segment rigidly connectedbetween steel and bronze segments. Find the largest safevalue of P if the working stresses are 120 MPa for steel,68MPa for aluminum, and110 MPa for bronze.

Q#3:The wood pole is supported by two cables of 0.25 in.diameter. The turnbuckles in the cables are tightened untilthe stress in the cables reaches 60000 psi. If the workingcompressive stress for wood is 200 psi, determine thesmallest permissible diameter of the pole.


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Q#4:Find the maximum allowable value of P for thecolumn. The cross-sectional areas and working stresses() are shown in the figure.

Q#5:Determine the smallest safe cross-sectional areas ofmembers CD, GD, and GF for the truss shown. Theworking stresses are 140 MPa in tension and 100 MPa in

compression.(The working stress in compression is smallerto reduce the danger of buckling.)

Q#6.Find the stresses in members BC, BD, and CF for thetruss shown.Indicate tension or compression.The cross-sectional area of each member is 1400 mm2.

Q#7:Determine the smallest allowable cross-sectionalareas of members BD, BE, and CE of the truss shown. The

working stresses are 20 000 psi in tension and 12000 psi incompression. (A reduced stress in compression is specifiedto reduce the danger of buckling.)

Engr. Khalid Yousaf BS Civil Engineering(The University of Lahore)

Q#8 Th if 300 lb b AB i 500 lb ti l


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Q#8: The uniform 300-lb bar AB carries a 500-lb verticalforce at A. The bar is supported by a pin at B and the 0.5 in.diameter cable CD. Find the stress in the cable.Q#9:Determine the smallest allowable cross-sectional areasof members CE, BE, and EF for the truss shown. Theworking stresses are 20 ksi in tension and 14 ksi incompression. (The working stress in compression is smaller

to reduce the danger of buckling.)Q#10:For the Pratt bridge truss and loading shown, determinethe average normal stress in member BE, knowing that thecross-sectional area of that member is 5.87 in2.

Q#11:A solid brass rod ABand a solid aluminum rod BCare connected together by a coupler at B, as shown inFigure Q#11.The diameters of the two segments are =60and = 50, respectively. Determine the axial

stresses(in rodAB) and(in rodBC).

Engr. Khalid Yousaf BS Civil Engineering The University

Q#12 A h ll i l l i ( Fi 1 23)


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Q#12:A hollow circular nylon pipe (see Fig 1-23) supportsa load= 7800,which is uniformly distributed arounda cap plate at the top of the lower pipe. A second load isapplied at the bottom. The inner and outer diameters of theupper and lower parts of the pipe are = 51, =60,= 57, and = 63, respectively. The

upper pipe has a length = 350; the lower pipelength is = 400 .Neglect the self weight of thepipes.

(a) Find so that the tensile stress in upper part is 14.5MPa. What is the resulting stress in the lower part?

(b) Ifremains unchanged, find the new value of sothat upper and lower parts have same tensile stress.

(c) Find the tensile strains in the upper and lower pipesegments for the loads in part (b) if the elongation of theupper pipe segment is known to be 3.56 mm and thedownward displacement of the bottom of the pipe is 7.63mm.

Q#13 A 1 i di t lid b (1) lid b


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Q#13:A 1- in. diameter solid bar (1), a square solid bar(2), and a circular tubular member with 0.2 in. wallthickness (3), each supports an axial tensile load of 5 kips.

(a) Determine the axial stress in bar (1).

(b) If the axial stress in each of the other bars is 6 ksi, what

is the dimension,b

, of the square bar, and what is the outerdiameter,c, of the tubular member?

Q#14:The Washington Monument (Figure Q#14.) stands555 ft. high and weighs 181,700 kips (i.e., approximately182 million pounds). The monument was made from over36,000 blocks of marble and granite. As shown in Fig. 1b,the base of the monument is a square that is 665.5 in. longon each side, and the stone walls at the base are 180 in.

thick.Determine the compressive stress that the foundationexerts over the cross section at the base of the monument,assuming that this normal stress is uniform.

Q#15:The pin-jointed planar truss in is subjected to a singledownward force Pat joint A.All members have a cross-sectionalarea of 500 mm2. The allowable stress in tension is ()allow300 MPa, while the allowable stress (magnitude) in compressionis(

)allow

200 MPa. Determine the allowable load,allow

.

Q#16 A l i t t b ildi i f b i t d f


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Q#16:A column in a two-story building is fabricated fromsquare structural tubing having the cross-sectionaldimensions shown in Fig b.Axial loads= 200 and= 200 are applied to the column at levelsAandB,as shown in Fig a. Determine the axial stress insegment AB of the column and the axial stress in

segment BC

of the column. Neglect the weight of thecolumn itself.

Q#17:Each member of the truss in Fig. is a solid circularrod with diameter = 10 . Determine the axial stress in the truss member (1) and the axial stress in thetruss member (6).

Q#18:The three-part axially loaded member in Fig.

consists of a tubular segment (1) with outer diameter()1= 1.00 . and inner diameter ()1= 0.75 . asolid circular rod segment (2) with diameter =1.00 ., and another solid circular rod segment (3) withdiameter= 0.75 . The line of action of each of thethree applied loads is along the centroidal axis of themember. Determine the axial stresses , , and ineach of the three respective segments.

MOS-I (Lecture 1-2) Engr. Khalid Yousaf

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