Chapter 7a-Part A
Transcript of Chapter 7a-Part A
CIVL222 STRENGTH OF MATERIALS
Chapter 7a-Part A
Stresses in Beams – Pure Bending
Instructor: Assoc.Prof.Dr. Mürüde Çelikağ
Bending• Today’s Objective:
• Students will be able to:
a) Determine the stress in a beam member caused by bending
In-class Activities:• Flexural formula• Unsymmetrical bending•Composite Beams• Concept Quiz
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Bending Deformation of a Straight Member
• When a bending moment is applied to a straight prismaticbeam, the longitudinal lines become curved and verticaltransverse lines remain straight and yet undergo a rotation
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Bending Deformation of a Straight Member
• A neutral surface is where longitudinal fibers of the materialwill not undergo a change in length.
• The bottom material is stretched and the top material iscompressed – leaving a neutral surface somewhere inbetween.
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Bending Deformation of a Straight Member
Assumptions:
a) Plane section remainsplane
b) Length of longitudinalaxis remains unchanged
c) Plane section remainsperpendicular to thelongitudinal axis
d) In-plane distortion ofsection is negligible
Before deformation
After deformation
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Bending Deformation of a Straight Member
Now lets look at a segment in isolation: x does not change inlength but x does.
xxx
x
0
lim
From the strain equation:
yx 0
lim
Substituting in terms of angle :
y
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Bending Deformation of a Straight Member
y
It should be noted that longitudinal normal strain varieslinearly with distance y from the neutral axis, ie:
yE
E
x
xx
Hooke’s Law
stress
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The Flexure FormulaThe stress resultants that are related to the normal stress xacting on the cross section are
IdAy 2
Then
EIM moment-curvature equation
Bending stress variation
dAyEydAyEdAyM
ydAEdAN
AAAx
AAx
2
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The Flexure Formula
Bending stress variation
But we also know that
yE
yE
1
Substitute into moment-curvature equation
yEEIM Solve for
IMy
Flexure Formula
EIM
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The Flexure FormulaProcedure for analysisInternal moment
• Section member at point where bending or normal stress isto be determined and obtain internal moment M at thesection.
• Centroidal or neutral axis for cross-section must be knownsince M is computed about this axis.
• If absolute maximum bending stress is to be determined,then draw moment diagram in order to determine themaximum moment in the diagram.
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The Flexure Formula
AyydAA
• The z and y axes must pass through the centroid of the cross section.• Therefore NEUTRAL AXIS Passes through the centroid of the
cross section
From statics
ii
iii
ii
iii
A
Ayy
A
Axz
~ ,
~
Location of Neutral Axis
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The Flexure Formula
Procedure for analysisSection property
• Determine moment of inertia I, of cross-sectional area aboutthe neutral axis.
• Methods used are discussed in Textbook Appendix A.
• Refer to the course book’s inside front cover for the values ofI for several common shapes.
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The Flexure FormulaMoment of Inertia of composite area
A composite area is made by adding or subtracting a series of“simple shaped areas” like
Rectangles:
Triangles:
Circles:
3
centroidalx-axis
;12
bhI
3
centroidalx-axis
;36
bhI
4
cen tro id al cen tro id a lx -ax is y-ax is 4
rI I du e to sym m etry
3
centroidaly-axis 12
hbI
3
centroidaly-axis 36
hbI
2xy'y AdII 2
yx'x AdII PARALLEL-AXISTHEOREM
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The Flexure Formula
Procedure for analysisNormal stress
• Specify distance y, measured perpendicular to neutral axisto point where normal stress is to be determined.
• Apply equation = -My/I, or if maximum bending stress isneeded, use max = Mc/I.
• Ensure units are consistent when substituting values into theequations
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Example #2A beam having a tee-shaped cross section is subjected to equal 18 kN.m bendingmoments. As shown in left figure. The cross-sectional dimensions of the beamare shown in right figure. Determine:(a) The centroid location, the moment of inertia about the z axis, and the
controlling section modulus about the z axis.(b) The bending stress at point H, state whether the normal stress at H is tension
or compression.(c) The maximum bending stress produced in the cross section. State whether
the stress is tension or compression.
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Unsymmetric BendingMoment applied along principal axis
If y and z are the principal axes. ∫yz dA = 0(The integral is called the product of inertia)
FR = Σ Fx; 0 = ∫ σ dAA
(MR)y = Σ My; 0 = ∫ z σ dAA
(MR)z = Σ Mz; 0 = ∫ – y σ dAA
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Unsymmetric Bending
= +
Alternatively, identify the orientation of the principal axes (ofwhich one is the neutral axis). Orientation of neutral axis:
tantany
z
IIa
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Unsymmetric Bending
Note: The angle α, which is measured positive from the +zaxis toward the +y axis, will lie between the line of actionof M and the y axis; ie θ ≤ α ≤ 90°.
= +
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Composite BeamsAPPLICATIONS
Reinforced concrete beam Sandwich panel
Fig. 6.38(b) Fig. 6.38(a)
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Composite Beams
Transformed homogeneous beam obtained through a transformation
factor:n = E1
E2
dF = σ dA = σ’ dA’
σ dz dy = σ’ n dz dy
σ = n σ’
and
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1) Provided that the bending formation of a straight member is small and within elastic range. Which of the following statements is incorrect?
A) Plane section remains plane C) The length of the longitudinal axis remains unchanged
B) Cross section remains perpendicular D) In-plane distortion of cross section is to the longitudinal axis not negligible
2) Which of the following statements is incorrect for bending of a straight member?
A) Bending stress is proportional to C) bending stress is inversely the moment proportional to the moment of
inertia of the section
B) Bending stress is inversely proportional D) bending stress is not a function to the second moment of area of the section of the location
READING QUIZ
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1) Which of the following statements is true?
The flexure formula for a straight member can be applied only
A)when bending occurs about axes that represent the principal axes of inertia for the section.
B) The principal axes have their origin at the centroid.C) The principal axes are orientated along an axis of symmetry, if there
is one, and perpendicular to it.D)All of the above.
CONCEPT QUIZ
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