ME 323 – Mechanics of Materials

Exam 2 ReviewFall 2019

Joshua Pribe

Exam Information

• Date: Wednesday, November 6

• Time: 8 – 10 PM

• Room: WALC 1055• See course website for seating arrangement

• Equation sheet is posted online; a hard copy will be provided during the exam

• In the exam, do NOT write on the back of the pages

• Format: 3 calculation problems; 1 multiple-part “conceptual” problem

2

Exam Coverage

• Lectures: 15 – 29

• Lecture Book: Chapters 10 – 11, 16 – 17

• Homework: 6 – 9

• Topics• Normal (flexural) and shear stress in beams

• Don’t forget how to draw shear force and bending moment diagrams – you may need these to find the correct V and M to use when calculating the stresses

• Beam deflections (second-order integration)

• Energy methods (Castigliano’s theorem)

• Finite element methods

3

Ch. 10: Beam stresses

4

Flexural stress (normal stress due to bending):

Transverse shear stress (shear stress due to bending):

Review material: HW 6, Quiz 6, Week 6 Pre-Week Videos, “Summary of Normal and Shear Stress in Beams”

x

My

I = −

xyIt

VQ =

Internal shear force and bending moment in beams

y

z

* *Q A y=

3

zz

4 4

zz

12

4 64

bh

r dI I

I I

= =

=

=

= (rectangle)

(circle)

Ch. 10: Stresses in beams

5

Asymmetric cross sections: is the normal stress magnitude the same at the top and bottom of the beam?Hint: Which is farther from the neutral axis?

Finding the shear stress at a point on the cross section:

From Practice Problems 2 on the course blog Lecture Book Example 10.7

Normal and shear stress distributions at point C:-Find the reaction forces, make a cut at C, use equilibrium to find MC and VC, and calculate the stresses

Maximum normal and shear stress in the beam:-Use shear force and bending moment diagrams to find maximum magnitudes of M(x) and V(x)

Ch. 11: Beam deflections

6

Moment-curvature equation (2nd-order integration): ''( ) ( )EIv x M x=

Review material: HW 7, Quiz 8, Week 7 Pre-Week Videos, summary at end of Lectures 20-23 slides

Integrate once to get the slopeIntegrate again to get the deflectionRemember integration constants; use boundary and continuity conditions to solve for unknowns

'( ) ( )v x x( )v x

M(x) > 0: compression in +y fibers of beam (beam is “smiling”)θ > 0, v’(x) > 0: CCW rotationv(x) > 0: upward deflection

x axis starts at left end of the beam

Boundary conditions and continuity conditions

7

BC type BCs for 2nd-order method

fixed

simple support (pin or roller)

Free end or concentrated force or moment

0' 0

vv==

0v =

none

0v =

P

M0

none

none

CC type CCs for 2nd-order method

roller

discontinuity in load function

point force

2 1

2 1

0' '

v vv v= ==

2 1

2 1' 'v vv v==

point moment

2 1

2 1' 'v vv v==

2 1

2 1' 'v vv v==

Ch. 11: Beam deflections

8

Determinate problems

Quiz 8: Split the beam into sections whenever the loading changes; use BCs and CCs, find deflection at a point

Lec. 20, Ex. 2: Use BCs to find integration constants; sketch deflection curve

Indeterminate problemsPractice problems 2 on course blog: find M(x)

in terms of the reaction at A; use BCs to solve for integration constants and the reaction at A

x

Ch. 11: Beam deflections – Superposition

9

From lecture 23 (see Prof. Gonzalez’s slides in 11:30 section) (see also HW8, P1)

=

+

0 0p w= −

BP R=

Ch. 16: Energy methods

10Review material: HW 8, Quiz 9, Week 9 Pre-Week Videos, summary at end of Lectures 24-26 slides

“Neglect shear strain energy due to bending”→ don’t include this term

Castigliano’s second theorem

1,2,...,

1, 2,...,

1, 2,...,

i P

i

i M

i

i T

i

Ui N

P

Ui N

M

Ui N

T

= =

= =

= =

Remember: δi is the displacement at the point where Pi is applied and in the direction of Pi

θi is the angle of rotation at the point where Mi is applied and in the direction of Mi

0 1,2,..., R

i

Ui N

R

= =

Ri are redundant reactions

Remember to add dummy loads when neededExpress the strain energy in terms of applied loads, dummy loads, and redundant reactionsSet the dummy loads to zero after differentiating

Ch. 16: Energy methods

11

Determinate problems

Quiz 9: Horizontal dummy loads at C and D to find horizontal displacements at C and D

Lecture Book Ex. 16.5:Find vertical deflection at D →Need to find reaction at C in terms of PSplit the beam into segments at changes in loading or geometryEasiest to keep the free end when you cut each segmentYou may use local coordinate axes for each segment

/U P

Indeterminate problems

Practice problems 2 on course blog: Find reaction at B and rotation angle at end A →apply a dummy moment at A and choose By as the redundant reaction

Ch. 17: Finite element methodsProcedure (remember: )

1. Define the nodes and elementsFor N elements, there are N+1 nodes

2. Construct the stiffness matrix [K]

a) [K] is symmetric and tri-diagonal

3. Construct the force vector {F}

a) This is the external force on each node

4. Enforce the displacement BCsa) For fixed displacement at node i, eliminate:

The ith row and column of [K], and the ith row of {F}

5. Solve for the unknown displacementsa) Use the displacements to find:

12

( ),avg i

i

i

Ak

E

L=

( )1i

i i i

i

Eu u

L + −=

{ } [ ]{ }F K u=

Stress in element i: Internal force in element i: ( ) ( ),

1 1

( )avg i

i i i i i i

i

EAu u uk u

LF + +− = −=

External reaction forces: use the crossed-out rows of the stiffness matrix (see Quiz 10)

Review material: HW 9, Quiz 10, Week 10 Pre-Week Videos

where

General reminders

• Consistent sets of units

• Practice the conceptual questions, additional examples, sample exams, homework problems, and quiz solutions posted on the course website!• Practicing lots of examples is the best way to study

• There are some problems you can solve using second-order integration or Castigliano’s theorem (two practice problems for the price of one!). Note that in the posted sample exams, we tell you which method to use

• If you are stuck on a problem, don’t be afraid to skip it and come back later13

Metric, option 1

Force N

Length m

Stress Pa

Metric, option 2

Force N

Length mm

Stress MPa(1 MPa = 1 N/mm2)

Remember: 1 MPa = 106 Pa, 1 GPa = 109 Pa

US units, option 1

Force lb

Length in

Stress psi

US units, option 1

Force kips

Length in

Stress ksi

Remember: 1 kip = 103 lb, 1 ksi = 103 psi