Mechanics of Materials (M.o.M)- Morning Session Approximate 7%-10% Approximate 15 questions~30 minutes MAX Review 18 questions (taken from Lindeburg’s book, 2009) Feb 24th, 2014 •

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Supplied Reference Handbook Page 76 - 82 in M.o.M session Approximate 5%-7% Page 56 -62 in Material Science/Structure of Matter Session  Approximate 2% READING =FREE pts Page 79 for more Column Design M.o.M Session,  Approximate 1%

02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

Stress and Strain-M.o.M Session  Get the equations from Page 76 M.o.M session  Given E=3x10^7 N/(cm^2) rod, L = 30cm, α temperature

coefficient of expansion=6x10^-6cm/(cm.◦C), A =2 cm^2  BC: Fixed at both ends Q 1 : If the rod is heated to 60 ◦C above the neutral temp, what is the nearly stress? (Ans unit: N/cm^2) a. 110 b. 11000 c. 36000 d. 57000  Thermal strain=α temperature coefficient of expansion* given T  Stress=E*thermal strain  b 2

02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

Stress and Strain, Thermal Expansion M.o.M Session  Get the equations from Page 76 in M.o.M session  Given E=3x10^7 N/(cm^2) rod, L = 30cm, α- temperature

coefficient of expansion=6x10^-6cm/(cm.◦C), A =2 cm^2 Q 2: If one end of the rod is free to expand (cantilever condition), the elongation is most nearly?(Ans unit: cm) a. 5.4 x 10^-4 b. 3.6 x 10^-4 c. 0.01 d. 0.03  Thermal strain=α- temperature coefficient of expansion* given T

 Elongation=thermal strain *L c

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02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

Stress and Strain-Uniaxial elongationM.o.M Session  Get the equations from Page 76 M.o.M session  Q 3: Given an aluminum pipe 2m long, uniformly carries a

a.

b. c. d.   

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196.2N axial tensile load. Outer diameter of the pipe is 200mm, inner diameter of the pipe is 105mm. Given that Young’s modulus for aluminum is 75GPa, the longitudinal deformation of the pipe in mm is most nearly: 0.0023 0.00074 0.00023 0.000172 A=(pi/4)*(do^2-di^2) E=75*10^9/(1000^2) (N/mm^2) Elongation=PL/EA c

02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

Stress and Strain - M.o.M Session  Get the equations from Page 77 in M.o.M session

Q 4:The maximum in-plane shear stress (ksi) in the element shown below is most nearly: 20ksi a. 10 b. 14.1 10ksi c. 44.1 40ksi d. 316

 in-plane shear stress (ksi)=R  R=sqrt(((40-20)/2)^2+10^2) b 5

02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

Cylindrical Pressure, and Torsional Stress-M.o.M Session  Get the equations from Page 78 in M.o.M session, J from page

66 table in Statics session Q 5: A 1020 carbon steel rod is 0.25cm diameter and 6 cm long. The shear modulus is 11.5 x 10^6 N/(cm^2). Most nearly what torque must be applied to twist the rod 8 degrees? (Ans unit: Ncm) a. 100 b. 120 c. 270 d. 420 J=(pi*r^4)/2 T=JG*phi /L keep in mind that phi is in radians (degree*2*pi*rad/360 deg)a

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02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

Torsional- Shear Stress-M.o.M Session  Get the equations from Page 78 in M.o.M session

Q 6: Given a beam with rectangular cross sectional area is subjected in pure bending, which of the following is the correct statement: a. Maximum shear stress occurs at the outermost fiber b. Maximum shear stress occurs at the neutral axis c. Constant shear stress occurs throughout the beam d. No shear stress is present throughout the beam  Transverse shear stress t= VQ/(Ib), V const., I const., b const., Q=A’y’ write Q in a function of y,  t=(V/2I)*(h^2/4-y^2)b

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02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

Cylindrical Pressure, and Torsional Stress-M.o.M Session  Get the equations from Page 77 in M.o.M session

Q 7: The pressure gage in an air cylinder reads 1680 kPa. The cylinder is constructed of a 12 mm rolled- steel plate with an internal diameter if 700mm. The tangential stress (MPa) inside the tank is most nearly: a. 25 b. 50 c. 77 d. 100 Check the ratio of t/r, if less than 1/10 Sigma t=Pr/t b

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02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

Beam-M.o.M Session  Read given equations page 78 in M.o.M session  Q 8: What is the maximum flexural stress at the distance x from a. b. c.

d.

the free end of a cantilever beam supporting a tip load P? Pxc/(2I) Pc/(2I) PcL/(2I) P Pxc/(2EI) L,I

c x

 Flexural stress=M(c/2)/I=Pxc/(2I)a

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02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

Beam/Uniaxial Deformation -M.o.M Session  Read given equations page 76 in M.o.M session  Q 9: What is most nearly the elongation in the cable if F=1000N, a. b. c.

d.

A=2cm^2, E=1.5x10^6N/(cm^2) (Ans unit: cm)? 0.0028 0.14 cable 0.28 3m 0.56 4m

 T cable=1000*5/3  Elongation=TL/AE c 10

02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

1000N

Beam/ Deformation due to bending-M.o.M Session 

Read given equations page 82 in M.o.M session

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Read the given Beam deflection Formula page 82 in M.o.M session

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Q 10: A 5m long beam is shown with fixed end and simply supported at the other end. Beam carries its own mass of 30.6kg/m. The modulus elasticity of the beam is 210GPa, the moment inertia is 2880cm^4. What is the reaction (N) at the simply supported end?

a.

72

b.

510

c.

560

30.6kg/m

d.

770

5m

A

B

Statically indeterminate, pick redundant By remove the support, replace with a force By upward  deflection caused from self weight + deflection caused from By has to be zerowL^4/(8EI)=PL^3/(3EI)

Change mass to weightw=30.6kg/m*(9.81m/s^2) P=3*wL/8 P=562.8N c 11

02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

Beam/ Deformation due to bending-M.o.M Session 

Read given equations page 78 in M.o.M session

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Read the given Beam deflection Formula page 39 in M.o.M session

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Q 11: Beam is shown with fixed end and simply supported at the other end. Beam carries its own mass of 30.6kg/m. The modulus elasticity of the beam is 210GPa, the moment inertia is 2880cm^4. The support at B is removed and replaced by a variable force. The force is adjusted until it is 900N. What is the net defection (mm) of the beam at 3.5m from the fixed end?

a.

0.29 downward

b.

0.32 downward

c.

-1.2 upward

d.

-1.9 upward

A

30.6kg/m

B

5m

You have solved RXTN By in Q 10, so now you know, 900N will cause a larger deflection in beam. Distance x is measured from fixed end=3.5m, a=L, deflection from 900N=(Px^2)(-x+3a)/(6EI)=-0.00349m, upward. Deflection from x position due to mass=0.00232m downward net deflection is 1.2mm c 12

02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

Columns-Bucking M.o.M Session  Read given equations page 79 in M.o.M session  Read given equations page 229 in Mechanical Engi session

 Given a fixed end column length of 3.5m with compressive

uniaxial load P applies from top of col., E=200x10^9 Pa, centroid moment of inertia is I=1.0x10^-4(m^4), given safety factor for buckling of 3.0  Q 12: What is the largest allowable axial load P? (Ans unit: kN) a. 1100 b. 1209 c. 1343 d. 1390 Pcr=(pi^2)EI/((L eff)^2), cantilever=1 fixed endL eff=2.1L, Pallow=Pcr/FSc

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02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

Columns-Buckling-M.o.M Session  Read given equations page 76 in M.o.M session  Q 13: A 10cm x 10cm square column supports a compressive

a. b. c. d. e.

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force of 9000N. The load is applied with an eccentricity of 2.5cm along one of the lines of symmetry, what is the maximum tensile stress in the column?(Ans unit kPa) 450 900 1350 2250 Calculate A of cross section, I of cross section, distance from the NA to the extreme fiber=b/2, Stress=P/A+ Pec/I, P is downward, neg.  450kPaa

02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

Strength of Materials  Read M.S. session page 56  Q 14: For a fixed curing time, the ultimate strength of concrete:

a. b. c. d.

Increase with decrease in water content Decrease with decrease in water content Is independent of water content is independent of curing pressure

Read the section for water content a

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02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

Strength of Materials  Read M.S. session page 58  Q 15: A micro crack is found inside a glass panel during the

a. b.

c. d. 

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beginning phase of a very light tensile loading test. The geometric parameter for the crack is : 0.5 0.7 1.1 1.0 d

02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

Strength of Materials  Read M.S. session page 59  Q 16: Which one below is the representative value of fracture a. b. c.

d. 

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toughness of alumina (ksi.in^0.5) 3.7 14.3 3.2 4.1  use the table d

02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

Composite section 

Read the given composite eqtn page 79 in M.o.M session

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Q 17: Beam is shown with simply supported at both ends. You can neglect beam own weight. The modular ratio is E steel/E timber=20. Distributed force is w=500lb/ft. Beam length is 18ft. Determine maximum normal stresses in steel and in the wood. Thickness for steel=0.375in, thickness for wood=7.5”, beam thickness=5.5”

a.

700psi for wood, 700 psi for steel

b.

620 psi for wood, 13.61ksi for steel

c.

13.61ksi for both

d.

620psi for steel, 620psi for wood

Recall weighted flexural rigidity: EI bar=Ew(Iw+nIs) b

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02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen

Stress-Strain diagram and bending stress  Read the given composite eqtn page 79 in M.o.M session  Q 18: A wide flange beam with overhang is shown. Determine the maximum

tensile stress on the cross section at point C. Neglect beam self weight. Sketch V and M diagrams, determine maximum flexural stress in beam.

a.

Maximum tensile stress at top fiber=5ksi, Maximum flexural stress=11ksi

b.

Maximum tensile stress at top fiber=11ksi, Maximum flexural stress=5ksi

Need: equilibrium in the whole beam By=1.25K, Dy=1.75K Recall sign convention, method of cutting in beam analysis. M at C=-2.5kft, Maximum tensile stress at top fiber= =-(-2.5kft*12*3in/(19.31416in4)=4.66ksi Take M max from the M dia, Flexural stress max =-Mmax *y/I

=11.14ksi a

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02/24/2014 FE Review- Mechanics of Materials- Lan Nguyen