engr 220 section 6.3 – 6.4. assumptions 1.straight prismatic beams 2. homogenous material 3. cross...

23
ENGR 220 Section 6.3 – 6.4

Post on 20-Dec-2015

239 views

Category:

Documents


1 download

TRANSCRIPT

Page 1: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

ENGR 220Section 6.3 – 6.4

Page 2: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

Assumptions

1.Straight prismatic beams

2. Homogenous material

3. Cross sectional area symmetric w.r.t an Axis

4. Bending moment applied perpendicular to this axis of symmetry

5. Neutral surface does not experience any change in length

6. All cross sections remain plane and perpendicular to the longitudinal axis.

7. Any deformation of the cross section in its own plane is neglected.

Page 3: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

Bending Deformation

Page 4: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

Bending Deformation

Page 5: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment
Page 6: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment
Page 7: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment
Page 8: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

Maximum Strain

max

max

max

c

y

c

y

c

Page 9: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

Strain / Stress Profiles

Page 10: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

Wood Specimen that failed in bending

Page 11: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

The Flexure Formula

• Applying Hooke’s Law– σ=Eε

• A linear variation in strain results in a linear variation in stress. max

c

y

max

c

y

Page 12: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

Locating Neutral Axis.

Resultant Normal Force on the Cross section = 0

dF = dA = 0

Substitute for , in terms of max

-(max /c ) y dA = 0

y dA = 0

y dA = y = location of centroid.

y = 0 implies the centroidal axes are lying on the neutral axis.

Page 13: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment
Page 14: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

Flexure Formula

I

Mcmax

Page 15: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

Moment of Inertia Review

• Note: Moment of inertia is actually a misnomer. It has been adopted because of its similarity to integrals of the same form related to mass.

xdACentroid

dAx2Inertia ofMoment Area

dmr 2Inertia ofMoment Mass

Page 16: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

Parallel Axis Theorem

xyy

yxx

AdII

AdII2

'

2'

Page 17: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

Composite Shapes

)(

)(2

'

2'

xyy

yxx

AdII

AdII

Page 18: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

Determine the area moment of inertia for the rectangle shown about the axis x’ and about the axis xb.

Page 19: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

Determine the internal moment M at the section caused by the stress distribution (1) using the flexure formula and (2) find the resultant of the stress distribution using basic principles

Page 20: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

Determine the Moment of Inertia about the z-axis for the following I-Beam.

Page 21: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

Determine the maximum bending stress at a-a

Page 22: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

The strut on the utility pole supports the cable having a weight of 600 lb. Determine the absolute maximum bending stress in the strut if A, B, and C are assumed pinned.

Page 23: ENGR 220 Section 6.3 – 6.4. Assumptions 1.Straight prismatic beams 2. Homogenous material 3. Cross sectional area symmetric w.r.t an Axis 4. Bending moment

The simply supported truss is subjected to the central distributed load. Neglect the effect of the diagonal lacing and determine the absolute maximum bending stress in the truss. Top member: 1” OD and 0.1875” wall thickness. Bottom member: 0.5” OD Solid Rod.