Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
Chapter 11 Columns
11.1 Introduction
Buckling Failures of Columns
1. Failures investigated so far in this course: failures caused by excessive s________ or d__________ strength and stiffness of members are important
2. Buckling failure of columns: long, slender
members loaded axially in c__________ deflects
l___________ b___________ may collapse
eventually (instead of failures by direct
compression of the material)
3. Example: compressing a plastic slender ruler,
stepping on an aluminum can, think plate of a
bridge under compression, etc.
4. Buckling is one of the major causes of failures in
structures should be considered in design process
11.2 Buckling and Stability
Idealized Structure to Investigate Buckling and Stability (βBuckling Modelβ)
1. Rigid bars π΄π΄π΄π΄ and π΄π΄π΅π΅ joined by a
pin connection ~ rotational spring
with stiffness π½π½π π is added at the pin
an idealized structure analogous
to the column structure shown
above (elasticity is concentrated vs
distributed)
2. Hookeβs law for the rotational spring
ππ = π½π½π π ππ
3. If the bars are perfectly aligned, the axial load ππ acts through the longitudinal line
spring is uns_________, and the bars are in direct c____________
4. Suppose point π΄π΄ moves a small distance laterally (by external disturbances, forces
or imperfect geometry) rigid bars rotate through small angles ππ
Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
5. Axial forces and βrestoring momentβ πππ΅π΅ developed in the spring show opposite
effects Axial force tends to __________ the lateral displacement, and πππ΅π΅ tends
to ____________ it
6. What happens after the disturbing force is removed?
1) Small ππ ππ keeps __________ returns to the original position: Stable
2) Large ππ ππ keeps __________ fails by lateral buckling: Unstable
βHow large ππ should be to make the system unstable?β Critical load
Critical Load ππππππ
1. Moment in the spring: πππ΅π΅ = 2π½π½π π ππ
2. Under small angle ππ, the lateral
displacement at point π΄π΄: ππππ/2
3. Moment equilibrium for bar π΄π΄π΅π΅
πππ΅π΅ β ππ β οΏ½ππππ2οΏ½ = 0
οΏ½2π½π½π π βππππ2οΏ½ β ππ = 0
4. First solution of equilibrium equation: ππ = 0 trivial solution representing the
equilibrium at perfectly straight alignment regardless of the magnitude of the load
5. Second solution of equilibrium equation:
ππππππ =4π½π½π π ππ
The structure is in equilibrium regardless of the magnitude of the angle ππ
Critical load is the only load for which the structure will be in equilibrium in the
disturbed position, i.e. ππ β 0
6. What if ππ β ππππππ, i.e. canβt sustain the equilibrium?
1) If ππ < ππππππ, restoring moment is dominant structure is __________
2) If ππ > ππππππ, effect of the axial force is dominant structure is __________
7. From the critical load derived above, it is seen that one can increase the stability by
___________ing stiffness or ___________ing length
Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
Summary
1. ππ = 0: no disturbance equilibrium for any ππ
2. Disturbance introduced to cause ΞΈ β 0 and the
source of the disturbance removed
1) ππ < ππππππ: goes back to the original equilibrium
(stable equilibrium)
2) ππ = ππππππ: can sustain the equilibrium regardless
of ΞΈ (neutral equilibrium) ~ βbifurcationβ point
3) ππ > ππππππ: cannot sustain the equilibrium (unstable equilibrium)
3. These are analogous to a ball placed upon a smooth surface
Example 11-1: Consider two idealized columns. The first one consists of a single rigid
bar π΄π΄π΄π΄π΅π΅π΄π΄ pinned at π΄π΄ and laterally supported at π΄π΄ by a spring with translational
stiffness π½π½. The second column consists of two rigid bars π΄π΄π΄π΄π΅π΅ and π΅π΅π΄π΄ that are joined
at π΅π΅ by an elastic connection with rotational stiffness π½π½π π = οΏ½25οΏ½π½π½ππ2. Find an expression
for critical load ππππππ for each column.
Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
11.3 Columns with Pinned Ends
Differential Equation for Deflection of an βIdeal Columnβ (i.e. perfectly straight) with
Pinned Ends
1. Bending-moment
equation:
πΈπΈπΈπΈπΈπΈβ²β² = ππ
2. Moment equilibrium
equation:
ππ + πππΈπΈ = 0
3. Therefore, the deflection
equation of the
deflection curve is
πΈπΈπΈπΈπΈπΈβ²β² + πππΈπΈ = 0
4. Homogeneous, linear, differential equation of second order with constant coefficients
Solution of Differential Equation
1. For convenience, we introduce ππ2 = ππ/πΈπΈπΈπΈ
2. Rewrite the differential equation:
πΈπΈβ²β² + ππ2πΈπΈ = 0
3. From mathematics, the general solution of the equation is
πΈπΈ = π΅π΅1 sinππππ + π΅π΅2 cosππππ
4. Boundary conditions to determine π΅π΅1 and π΅π΅2:
πΈπΈ(0) = πΈπΈ(ππ) =
5. From the first condition, π΅π΅2 =
6. Thus the deflection of the column is πΈπΈ(ππ) = π΅π΅1 sinππππ
7. From the second condition, π΅π΅1 sinππππ =
8. Case 1: π΅π΅1 = πΈπΈ(ππ) = , i.e. the column remains straight (for any ππππ)
9. Case 2: sinππππ = βBuckling equationβ
Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
The column sustains equilibrium if ππππ = ππππ, ππ = 1,2,3, β¦
The corresponding axial (critical) loads are
ππ =ππ2ππ2πΈπΈπΈπΈππ2
10. Deflection curves at neutral equilibrium at critical loads are
πΈπΈ(ππ) = π΅π΅1 sinππππ = π΅π΅1 sinππππππππ
Critical Loads
1. The lowest critical load for a column with pinned ends:
ππππππ =ππ2πΈπΈπΈπΈππ2
2. The corresponding buckled shape (mode shape):
πΈπΈ(ππ) = π΅π΅1 sinππππππ
3. Note: the amplitude π΅π΅1 of the buckled shape is un____________ (but small)
4. ππ = 1: βFundamentalβ buckling mode
5. As ππ increases, βhigher modesβ appear
No practical interest because the fundamental load is reached first
To make higher modes occur, lateral supports should be provided at intermediate
points
Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
11.1 Columns with Pinned Ends (continued)
Critical Stress
1. Critical stress: the stress in the column when ππ =
ππcr =ππcrπ΄π΄
=ππ2πΈπΈπΈπΈπ΄π΄πΏπΏ2
2. Using the radius of gyration ππ = οΏ½πΈπΈ/π΄π΄
ππcr =ππ2πΈπΈ
(πΏπΏ/ππ)2
where πΏπΏ/ππ is called βslenderness ratioβ
3. Eulerβs curve: critical stress versus the
slenderness ratio
- Long and slender columns: buckle at
_______ stress
- Short and stubby columns: buckle at
_________ stress
- The curve is valid only for Ο < Οpl because we use ___________βs law
Effects of Large Deflections, Imperfections, and Inelastic Behavior
1. Ideal elastic column with small defections (Curve
A): No deflection or undetermined deflection at ππ =
ππcr
2. Ideal elastic column with large deflection (Curve B):
Should use exact (nonlinear) expression for the
curvature, i.e. instead of π£π£β²β² Once the column
begins buckling, an increasing load is required to
cause an increase in the deflections
3. Elastic column with imperfections (Curve C): imperfections such as initial curvature
imperfections produce deflections from the onset of loading; the larger the
imperfections, the further curve C moves to the right
4. Inelastic column with imperfections (Curve D): As the material reaches the
proportional limit, it becomes easier to increase deflections
Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
Example 11-2: A long, slender column ABC is pin
supported at the ends and compressed by an
axial load ππ. Lateral support is provided at the
midpoint B (only in the direction within the plane).
The column is constructed of a standard steel
shape (IPN 220; Table E-2) having πΈπΈ = 200 GPa
and proportional limit ππpl = 300 MPa. The total
length πΏπΏ = 8 m. Determine the allowable load
ππallow using a factor of safety ππ = 2.5 with
respect to Euler buckling of the column.
Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
11.4 Columns with Other Support Conditions
Column Fixed at the Base and Free at the Top
1. Bending moment at distance π₯π₯ from the base is
ππ = ππ(πΏπΏ β π£π£)
2. Bending moment equation: πΈπΈπΈπΈπ£π£β²β² = ππ =
3. Using the notation ππ2 = ππ/πΈπΈπΈπΈ again, the equation
becomes
π£π£β²β² + ππ2π£π£ = ππ2πΏπΏ
4. Homogeneous solution (the same as the pinned-pinned case):
π£π£H = πΆπΆ1 sinπππ₯π₯ + πΆπΆ2 cosπππ₯π₯
5. Particular solution:
π£π£P =
6. Consequently, the general solution is
π£π£(π₯π₯) = π£π£H + π£π£P = πΆπΆ1 sinπππ₯π₯ + πΆπΆ2 cosπππ₯π₯ +
7. Boundary conditions: π£π£(0) = , π£π£β²(0) = and π£π£(πΏπΏ) =
8. From the first condition, πΆπΆ2 =
9. From the second boundary condition, πΆπΆ1 =
10. Finally, the solution is π£π£(π₯π₯) = πΏπΏ(1 β cosπππ₯π₯) shape is identified but the amplitude
is und___________
11. From the third boundary condition, Ξ΄ cosπππΏπΏ =
12. The nontrivial solution (i.e. buckling equation) is
cosπππΏπΏ = 0
13. Therefore, πππΏπΏ = ππΟ2
, ππ = 1,3,5, β¦
14. The critical loads are
ππcr =ππ2ππ2πΈπΈπΈπΈ
4πΏπΏ2, ππ = 1,3,5, β¦ and for ππ = 1,ππcr =
ππ2πΈπΈπΈπΈ4πΏπΏ2
Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
15. Buckled mode shapes are π£π£(π₯π₯) = πΏπΏ οΏ½1 β cos ππππππ2πΏπΏοΏ½
Effective Lengths of Columns
1. Effective length of a column: the length of the
equivalent pinned-end column having a deflection
curve matching the deflection of the given column
2. As seen in the figure, the effective length of a
column fixed at the base and free at the top is
πΏπΏππ =
3. From the critical loads of the two column cases, we
can derive a general formula for the critical load,
ππcr =ππ2πΈπΈπΈπΈπΏπΏππ2
Column with Both Ends Fixed against Rotation
1. According to the deflected
shape sketched based on the
boundary conditions, it is noted
that πΏπΏππ =
2. Therefore, the critical load is
ππcr =4ππ2πΈπΈπΈπΈπΏπΏ2
Column Fixed at the Base and
Pinned at the Top
1. By solving the differential
equation (details in the
textbook), we find the buckling
equation
πππΏπΏ = tanπππΏπΏ
2. Solving this equation
numerically, πππΏπΏ = 4.4934
3. The corresponding critical load is Pcr = 20.19πΈπΈπΈπΈπΏπΏ2
= 2.046ππ2πΈπΈπΈπΈπΏπΏ2
Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
4. The effective length is πΏπΏππ = 0.699πΏπΏ β 0.7πΏπΏ
Summary of Results
Example 11-3: A viewing platform
is supported by a row of
aluminum pipe columns having
length πΏπΏ = 3.25 m and outer
diameter ππ = 100 mm. Because
of the manner in which the
columns are constructed, we
model each column as a fixed-
pinned column. The columns are
being designed to support
compressive loads ππ = 200 kN. Determine the minimum required thickness π‘π‘ of the
columns if a factor of safety ππ = 3 is required with respect to Euler buckling. The
modulus of Elasticity of the aluminum is πΈπΈ = 72 GPa and the proportional stress limit is
480 MPa.
Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
(Intended Blank for Notes)
Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
11.5 Columns with Eccentric Axial Loads
Differential Equation of Columns with Eccentricity
1. Consider a column with a small
eccentricity ππ under axial load ππ
2. This is equivalent to a column under
centric load ππ but with additional couples
ππ0 =
3. Bending moment in the column is
obtained from a free-body-diagram from
the moment equilibrium (around π΄π΄)
ππ = ππ0 + ππ(βπ£π£) = ππππ β πππ£π£
4. Differential equation
πΈπΈπΈπΈπ£π£β²β² = ππ = ππππ β πππ£π£
π£π£β²β² + ππ2π£π£ =
5. The general solution: π£π£ = πΆπΆ1 sinππππ + πΆπΆ2 cosππππ + ππ
6. Boundary conditions: π£π£(0) = π£π£(πΏπΏ) =
7. These conditions yield
πΆπΆ2 =
πΆπΆ1 = βππ(1 β cosπππΏπΏ)
sinπππΏπΏ= βππ tan
πππΏπΏ2
8. The equation of the deflection curve is
π£π£(ππ) = βππ οΏ½tanπππΏπΏ2
sinππππ + cosππππ β 1οΏ½
Note: the deflection for the centric load was π£π£(ππ) = πΆπΆ1 sinππππ = πΆπΆ1 sin πππππππΏπΏ
Undetermined (centric load) vs determined (eccentric load)
9. Critical load (both pinned ends)
ππcr =ππ2πΈπΈπΈπΈπΏπΏ2
Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
Maximum Deflection
1. Maximum deflection Ξ΄ occurs at the midpoint, thus
πΏπΏ = βπ£π£ οΏ½πΏπΏ2οΏ½
= ππ οΏ½tanπππΏπΏ2
sinπππΏπΏ2
+ cosπππΏπΏ2β 1οΏ½
= ππ οΏ½secπππΏπΏ2β 1οΏ½
2. Consider an alternative expression for ππ
ππ = οΏ½πππΈπΈπΈπΈ
= οΏ½ππππ2
ππcrπΏπΏ2=πππΏπΏοΏ½ππππcr
3. Using this, the maximum deflection is described in terms of the ratio ππ/ππcr
πΏπΏ = ππ οΏ½secοΏ½ππ2οΏ½ππππcrοΏ½ β 1οΏ½
4. Load-deflection diagram (β)
- The deflection increases as the load ππ increases, but nonlinear even if linear
elastic material is used s___________ rule does not work
- When the imperfection is increased from ππ1 to ππ2: the maximum deflection
increases by
- As the load ππ approaches the critical load ππcr the deflection increases without
limit
- An ideal column with a centrally applied load (ππ = 0) is the limiting case of a
column with an eccentric load (e > 0)
Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
Maximum Bending Moment
1. Maximum bending moment occurs when π£π£ =
ππmax = ππ(ππ+ )
2. Thus the maximum bending moment is
ππmax = ππππ secπππΏπΏ2
= ππππ secοΏ½ππ2οΏ½ππππcrοΏ½
Example 11-4: A brass bar π΄π΄π΄π΄ projecting
from the side of a large machine is loaded
at end π΄π΄ by a force ππ = 7 kN with an
eccentricity ππ = 11 mm. The bar has a
rectangular cross section with height β =
30 mm and width ππ = 15 mm. What is the
longest permissible length πΏπΏmax of the bar
if the deflection at the end is limited to 3 mm? For the brass, use πΈπΈ = 100 GPa.
Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
11.6 Secant Formula for Columns
Maximum Stress in a Column under an Eccentric Load
1. Maximum stress occurs at the (concave/convex) side of the column
ππmax =πππ΄π΄
+ππmaxπππΈπΈ
2. Maximum moment
ππmax = ππππ secοΏ½ππ2οΏ½ππππcrοΏ½
3. From ππcr = ππ2πΈπΈπΈπΈ/πΏπΏ2 and πΈπΈ = π΄π΄π΄π΄2 where π΄π΄ is the radius of gyration, the maximum
moment is described as
ππmax = ππππ secοΏ½πΏπΏ
2π΄π΄οΏ½ πππΈπΈπ΄π΄
οΏ½
4. Substituting this into the maximum
stress formula above,
ππmax =πππ΄π΄
+πππππππΈπΈ
secοΏ½πΏπΏ
2π΄π΄οΏ½ πππΈπΈπ΄π΄
οΏ½
=πππ΄π΄ οΏ½
1 +πππππ΄π΄2
secοΏ½πΏπΏ
2π΄π΄οΏ½ πππΈπΈπ΄π΄
οΏ½οΏ½
5. This so-called βsecant formulaβ
describes the maximum compressive stress in a column under eccentric load in
terms of πΈπΈ,ππ/π΄π΄ , πΏπΏ/π΄π΄ (s__________ ratio) and ππππ/π΄π΄2 (ecc____________ ratio)
6. For given ππmax and πΈπΈ, one can find the possible pairs of ππ/π΄π΄ and πΏπΏ/π΄π΄ for each
eccentricity level (ππππ/π΄π΄2) and plot a graph of secant formula (β)
7. For centric load, i.e. ππππ/π΄π΄2 = , the critical stress is
ππcr =ππcrπ΄π΄
=ππ2πΈπΈπΈπΈπ΄π΄πΏπΏ2
=ππ2πΈπΈ
(πΏπΏ/π΄π΄)2
8. Secant formula and graph let us know the load-carrying capacity of a column in
terms of slenderness and eccentricity (Trend?)
Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
Example 11-5: A steel wide-flange column of HE 320A
shape is pin-supported at the ends and has a length of
7.5 m. The column supports a centrally applied load ππ1 =
1800 kN and an eccentrically applied load ππ2 = 200 kN.
Bending takes place about axis 1-1 of the cross section,
and the eccentric load acts on axis 2-2 at a distance of
400 mm from the centroid C.
(a) Using the secant formula, and assuming πΈπΈ =
210 GPa, calculate the maximum compressive stress
in the column.
(b) If the yield stress for the steel is ππππ = 300 MPa, what
is the factor of safety with respect to yielding?
Dept. of Civil and Environmental Engineering, Seoul National University Junho Song 457.201 Mechanics of Materials and Lab. [email protected]
(Intended Blank for Notes)
Many thanks for your hard work in this semester to learn Mechanics of Materials. I wish you the very best on your remaining course work and future career and life.
Cheers, Junho