Chapter 4.pptx

4-3
Symmetric Member in Pure Bending
• Internal forces in any cross section are equivalent to a couple. The moment of the couple is the section bending moment.
• From statics, a couple M consists of two equal and opposite forces.
• The sum of the components of the forces in any direction is zero. • The moment is the same about any axis perpendicular to the plane of the couple and zero about any axis contained in the plane. • These requirements may be applied to the sums of the components and moments of the statically indeterminate elementary internal forces.


• Apply the elastic flexural formula to find the maximum tensile and compressive stresses.
m
Mc I
A cast-iron machine part is acted upon by a 3 kN-m couple. Knowing E = 165 GPa and neglecting the effects of fillets, determine (a) the maximum tensile and compressive stresses, (b) the radius of curvature.
3
1 90 20 3 1800 12 2 1 30 40 3 1200 18 2 12 12






I 868 103 mm 868 10 -9 m 4
4 - 17
Sample Problem 4.2
• Apply the elastic flexural formula to find the maximum tensile and compressive stresses.
Chapter 4 Pure Bending
The normal stresses and the curvature resulting from cure bending will be determined in this chapter
Pure Bending
Pure Bending: Prismatic members subjected to equal and opposite couples acting in the same longitudinal plane
L y
L L y y x m
L c y

ρ
y

c
(strain va ries linearly)

y c
or
m
x m
4-5
Stress Due to Bending (Elastic Flexure Formula)
3 1 I 12 bh 3 1 S 1 bh 6 Ah 6 c h2
Between two beams with the same cross sectional area, the beam with the greater depth will be more effective in resisting bending. • Structural steel beams are designed to have a large section modulus.
4 - 15
• Calculate the curvature
1


M EI
Parallel Axis Theorem: Mass moment of inertia
Suppose a body of mass m is made to rotate about an axis z passing through the body's center of mass. The body has a moment of inertia Icm with respect to this axis. The parallel axis theorem states that if the body is made to rotate instead about a new axis z′ which is parallel to the first axis and displaced from it by a distance d, then the moment of inertia I with respect to the new axis is related to Icm by
d is the perpendicular distance between the axes z and z′.
Sample Problem 4.2
SOLUTION:
Based on the cross section geometry, calculate the location of the section centroid and moment of inertia.
two materials with E1 and E2. • Normal strain varies linearly.
x
y

E1 y
• Piecewise linear normal stress variation.
1 E1 x 2 E2 x
quantified by the curvature of the neutral surface
1 Mc m m c Ec Ec I M EI
1
• Although cross sectional planes remain planar when subjected to bending moments, in-plane deformations are nonzero,
4-7
Properties of American Standard Shapes
4-8
Example 4.01
Example 4.02
Deformations in a Transverse Cross Section • Deformation due to bending moment M is
y x y z x y
• Expansion above the neutral surface and contraction below it cause an in-plane curvature,
1 anticlasti c curvature
Other Loading Types
• Eccentric Loading: Axial loading which does not pass through section centroid produces internal forces equivalent to an axial force and a couple • Transverse Loading: Concentrated or distributed transverse load produces internal forces equivalent to a shear force and a couple
• For static equilibrium,
y M y x dA c
First moment with respect to neutral plane is zero. Therefore, the neutral surface must pass through the section centroid.
• For a linearly elastic material,
x E x E m
y m (stress varies linearly) c y c
• For static equilibrium,
y Fx 0 x dA m dA c
m
Mc M I S I sectionmomentof inertia I sectionmodulus c S
A beam section with a larger section modulus will have a lower maximum stress
• Consider a rectangular beam cross section,
1


M EI
165 GPa 868 10 -9 m4
3 kN m
1
47.7 m
20.95 10 3 m-1
4 - 18
Bending of Members Made of Several Materials • Consider a composite beam formed from
Mc I M c A 3 kN m 0.022 m A I 868 10 9 mm 4 M cB 3 kN m 0.038 m B I 868 10 9 mm 4