020614 - Session

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Transcript of 020614 - Session

  • P.E. Civil Exam Review: Strength of Materials

    J.P. Mohsen

    Email:jpm@louisville.edu

  • Strength of Materials

    Stress vs. Strain

    Stress = Force / Area

    Strain = Change in Length / Original Length

    AP

    =

    LL

    =

    2

  • Stress Strain Diagram Typical for Ductile Materials

    Stre

    ss

    Strain

    3

  • Stress Strain Diagram

    Stre

    ss

    Strain

    Proportional limit

    4

  • Stress Strain Diagram

    Stre

    ss

    Strain

    Proportional limit

    Elastic limit

    5

  • Stress Strain Diagram

    Stre

    ss

    Strain

    Proportional limit

    Yield point

    Elastic limit

    6

  • Stress Strain Diagram

    Stre

    ss

    Strain

    7

  • Stress Strain Diagram

    Stre

    ss

    Strain

    Proportional limit

    Yield point

    Elastic limit

    Ultimate strength

    8

  • Stress Strain Diagram

    Stre

    ss

    Strain

    Proportional limit

    Yield point

    Elastic limit

    Ultimate strength

    Rupture

    9

  • Stress- Strain Relationship

    Stre

    ss

    Strain

    Proportional limit

    Yield point

    Elastic limit

    Ultimate strength

    Rupture

    True rupture

    10

  • Stress Strain Relationship Typical for Ductile Materials

    Stre

    ss

    Strain

    Hookes Law E=

    ExyportionlineartheofSlope =

    =

    =

    11

  • Areas under stress-strain curves: (a) modulus of resilience, (b) toughness, and (c) high-strength and high-toughness materials.

    Stre

    ss

    (a)

    Strain

    Modulus of resilience

    Stre

    ss

    (b)

    Strain

    Stre

    ss

    (c)

    Strain

    High strength

    Toughness

    Fracture

    High toughness

    12

  • Methods for estimating yield stress: (a) offset method and (b) extension method

    Stre

    ss

    (a)

    Strain, %

    Proportional limit

    0.2% offset yield strength

    Elastic limit

    0.2%

    Stre

    ss

    (b)

    Strain, %

    0.5% extension yield strength

    0.5%

    13

  • Problem 1)

    For the beam shown, compute the maximum bending stress in compression and maximum bending stress in tension. Also compute the shear stress at the neutral axis as well as at a section 6 in. from the bottom of the cross-section. Calculate the shear stresses at a point along the beam where the maximum shear force occurs.

    1500 lb

    1 in

    8 in

    4 in

    3.5 in 1 in

    INA = 97.0 in4

    NA

    1500 lb 4500 lb

    4 ft 4 ft 6 ft 6 ft

    14

  • Problem 1 (continued)

    For the beam shown, compute the shear stress at the NA and at 6 in. from the bottom of the cross section at the point where the maximum shear force occurs.

    1500 lb

    1 in

    8 in

    4 in

    3.5 in 1 in

    INA = 97.0 in4

    NA

    1500 lb 4500 lb

    4 ft 4 ft 6 ft 6 ft

    -1500 lb

    1500 lb

    3750 lb 3750 lb

    V (Ib)

    2250

    - 2250 - 2250

    1500

    15

  • Problem 1 (continued) - For the beam shown, compute the max. bending stress in tension and in compression.

    1500 lb

    1 in

    8 in

    4 in

    3.5 in 1 in

    INA = 97.0 in4

    NA

    1500 lb 4500 lb

    4 ft 4 ft 6 ft 6 ft

    -6000 -6000

    +7500

    M (ft-lb)

    ------ ---i---

    16

  • Bending Stress in beams :

    In the negative moment region In the positive moment region

    inCc 5.3= inCc 5.5=

    inCt 5.5= inCt 5.3=

    psit 324797)5.3)(12)(7500(==psit 408297

    )5.5)(12)(6000(==

    psic 510397)5.5)(12)(7500(==psic 259897

    )5.3)(12)(6000(==

    ICM c

    c =

    ICM

    =

    ICM t

    t =

    17

  • CISModulusSection =

    ICM

    = =

    SM

    36.175.5

    97 inCIS ===

    18

  • Problem 1 (continued)

    For the beam shown, compute the shear stress at the NA and at 6 in. from the bottom of the cross section at the point where the maximum shear force occurs.

    1500 lb

    1 in

    8 in

    4 in

    3.5 in 1 in

    INA = 97.0 in4

    NA

    1500 lb 4500 lb

    4 ft 4 ft 6 ft 6 ft

    -1500 lb

    1500 lb

    3750 lb 3750 lb

    V (Ib)

    2250

    - 2250 - 2250

    1500

    19

  • Shear stress in bending bIQV

    =3

    sec6 12]4[)1()3( inQ tioncrossofbottomfromin ==

    312.15]75.2[)1()5.5( inQ axisneturalat ==

    psitioncrossofbottomfromin 3.278)1)(97()12()2250(

    sec6 ==

    psiaxisneturalat 7.350)1)(97()12.15()2250(==

    20

  • What is the value of Q if shear stress is being calculated at:

    1 in

    8 in

    4 in

    3.5 in 1 in

    INA = 97.0 in4

    NA NA

    3 in 4.0 in

    NA 2.75 in 2.5 in

    5.5 in

    312]4[)1()3( inQ ==

    Neutral axis a section 6 in. above the base

    312.15]75.2[)1()5.5( inQ ==

    21

  • Problem 2.

    The vertical shear force acting on the I-section shown is 100 kN. Compute the maximum shear stress in bending acting on the this beam section.

    20 mm

    20 mm

    20 mm

    160 mm

    120 mm

    22

  • 20 mm

    20 mm

    20 mm

    160 mm

    120 mm

    NA

    TheoremAxisParallelAdbhI 23

    12+=

    2

    3

    1

    423

    1 000,520,19]90)[20)(120(12)20)(120( mmI NAabout =+=

    43

    2 667,826,612)160)(20( mmI NAabout ==

    423

    3 000,520,19]90)[20)(120(12)20)(120( mmI NAabout =+=

    4666,866,45 mmI NAabouttotal =20 mm

    23

  • Problem 2 (continued) Maximum shearing stress occurs at the neutral axis.

    3000,280]90)[20)(120(]40)[80)(20( mmNAatQ =+=

    NAatmmb 20=

    KNV 100=

    4666,866,45 mmI NAabouttotal =

    2/03.0)20()45866666(

    )280000()100( mmKN==

    bIQV

    =

    24

  • Problem 3.

    The simply supported beam has the T-shaped cross section shown. Determine the values and locations of the maximum tensile and compressive bending stresses.

    400 lb/ft 1000 lb

    A

    B

    D

    10 ft 4 ft

    y

    x

    RA = 1600 lb RB = 3400 lb

    6 in

    8 in

    0.8 in

    0.8 in

    25

  • Problem 3 (continued) Shear Diagram

    400 lb/ft 1000 lb

    A

    B

    D 10 ft 4 ft

    y

    x

    RA = 1600 lb RB = 3400 lb

    1600 1000

    -2400

    4 ft V (lb)

    26

  • Moment Diagram 400 lb/ft

    1000 lb

    A

    B

    D 10 ft 4 ft x

    RA = 1600 lb RB = 3400 lb

    1600 1000

    -2400

    4 ft V (lb)

    M (lb-ft) 4 ft

    -4000

    3200

    2

    27

  • Find y-bar.

    21

    2211

    AAyAyAbary

    ++

    =

    ( ) ( )8.44.6

    4.88.444.6++

    =

    in886.5=

    6 in

    0.8 in

    8 in

    A2

    A1 y1

    y2

    y-bar

    N.A.

    C

    0.8 in

    28

  • Compute Moment of Inertia, I.

    6 in

    0.8 in

    8 in

    A2

    A1 y1

    y2

    y-bar

    N.A.

    C

    0.8 in

    ( ) ( ) ( ) ( )

    ++

    += 2

    32

    3

    886.54.88.412

    8.06886.544.612

    88.0I

    449.87 inI =29

  • Stresses at x = 4 ft

    8.8 in

    y1 y-bar

    N.A. C

    cbot = 5.886 in.

    ctop = 2.914 in.

    4 ft

    ( )( ) psiI

    Mcbotbot 258049.87

    886.5123200=

    ==

    3200

    -4000

    ( )( ) psiI

    Mctoptop 127949.87

    914.2123200=

    ==

    ---i---

    Compression on top

    Tension in bottom 30

  • Stresses at x = 10 ft

    8.8 in

    y1 y-bar

    N.A. C

    cbot = 5.886 in.

    ctop = 2.914 in.

    10 ft

    3200

    -4000

    ( )( ) psiI

    Mcbotbot 323049.87

    886.5124000=

    ==

    ( )( ) psiI

    Mctoptop 159949.87

    914.2124000=

    ==

    ------

    Tension on top

    Compression in bottom 31

  • Problem 3 results.

    Identify maximum tensile and compressive stresses in the beam.

    ( ) psiT 2580max =

    ( ) psiC 3230max =

    (bottom of the section at x = 4 ft)

    (bottom of the section at x = 10 ft)

    32

  • Original State of Stress

    y

    x

    a

    a

    x

    y

    xy

    xy

    x

    y

    33

  • x

    y

    R

    x

    y

    xy

    xy

    y

    x x

    y

    xyConstruction of Mohrs circle from given stress components.

    34

  • Convention for plotting shear stress on Mohrs circle.

    Shear plotted up Shear plotted down

    35

  • y

    x

    y

    yx

    xy

    y

    x x

    y

    y

    yx y

    x

    x

    2

    1

    2 1

    1

    y

    1

    y

    a

    2

    1 2

    x

    xy

    b

    x

    x

    36

  • Problem 4 )

    For the state of stress shown, determine (a) the principal stresses; and (b) the maximum in-plane shear stress. Show the results on properly oriented elements.

    4 ksi

    8 ksi

    6 ksi

    y

    x

    37

  • (y-axis)

    (x-axis)

    (4,-6)

    (-8, 6)

    4 ksi

    8 ksi

    6 ksi

    y

    x

    y

    x

    Problem 4 (continued)

    38

  • R

    4 ksi

    8 ksi

    6 ksi

    y

    x

    (-8, 6)

    (4,-6)

    (y-axis)

    (x-axis)