Here are the steps to solve this example: 1) Find the moment of inertia, I, of the composite section about the neutral axis: I = I1 + I2 + A1*d1^2 + A2*d2^2 Where: I1 = moment of inertia of rectangular section 1 = b1*h1^3/12 = 4*6^3/12 = 144 in^4 I2 = moment of inertia of rectangular section 2 = b2*h2^3/12 = 2*4^3/12 = 32 in^4 A1 = area of section 1 = b1*h1 = 4*6 = 24 in^2 A

1. Chapter 1- Static engineering systems
1.1 Simply supported beams
1.1.1 determination of shear force
1.1.2 bending moment and stress due to bending
1.1.3 radius of curvature in simply supported beams subjected to
concentrated and uniformly distributed loads
1.1.4 eccentric loading of columns
1.1.5 stress distribution
1.1.6 middle third rule
1.2 Beams and columns
1.2.1 elastic section modulus for beams
1.2.2 standard section tables for rolled steel beams
1.2.3 selection of standard sections (eg slenderness ratio for
compression members, standard section and allowable
stress tables for rolled steel columns, selection of standard
sections)
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2. Stresses in beams
• Stresses in the beam are functions of x and y
• If we were to cut a beam at a point x, we would find a distribution of
direct stresses σ(y) and shear stresses σxy(y)
• Summing these individual moments over the area of the cross-section is
the definition of the moment resultant M,
• Summing the shear stresses on the cross-section is the definition of the
shear resultant V,
• The sum of all direct stresses acting on the cross-section is known as N,
2

3. • Direct stress distribution in the beam due to bending
• Note that the bending stress in beam theory is linear
through the beam thickness. The maximum bending
stress occurs at the point furthest away from the neutral
axis, y = c
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4. Flexure formula
•
Stresses calculated from the flexure formula are called bending
stresses or flexural stresses.
• The maximum tensile and compressive bending stresses occur at
points (c1 and c2) furthest from the neutral surface
• where S1 and S2 are called section moduli (units: in3, m3) of the cross-
sectional area. Section moduli are commonly listed in design
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handbooks

5. Euler’s Formula for Pin-Ended Beams
v v
v
v v
l
Putting
l v
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6. 6

7. 7

8. 8

9. Design of columns under centric loads
• Experimental data demonstrate
- for large Le/k σcr follows
le /r,
(le/k)2 Euler’s formula and depends
upon E but not σY.
- for small L/k σcr is
le e/r,
determined by the yield
strength σY and not E.
- for intermediate Le/k σcr
le /r,
depends on both σY and E.
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10. • For Le/r > Cc
l e/k
Structural Steel
π 2E σ
σ cr = σ all = cr
American Inst. of Steel Construction ( Le/kr ) 2 FS
l /
FS = 1.92
l e/k
• For Le/r > Cc
 ( Le /kr ) 2 
le / σ
σ cr = σ Y 1 − 2 
σ all = cr

 2Cc   FS
3
5 3 Le/kr 1  Le/k 
l / l /r
FS = + e −  e 
3 8 Cc 8  Cc 
 
le/k • At Le/k = Cc
le /r
2
2 2π E
σ cr = 1 σ Y Cc =
2 σY
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11. Sample problem
SOLUTION:
• With the diameter unknown, the
slenderness ration can not be evaluated.
Must make an assumption on which
slenderness ratio regime to utilize.
• Calculate required diameter for
assumed slenderness ratio regime.
• Evaluate slenderness ratio and verify
initial assumption. Repeat if
Using the aluminum alloy2014-T6, necessary.
determine the smallest diameter rod
which can be used to support the centric
load P = 60 kN if a) L = 750 mm,
b) L = 300 mm
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12. • For L = 750 mm, assume L/r > 55
• Determine cylinder radius:
P 372 × 103 MPa
σ all = =
A ( L r)2
60 × 103 N 372 × 103 MPa
2
= 2
c = 18.44 mm
πc  0.750 m 
 
 c/2 
• Check slenderness ratio assumption:
c = cylinder radius
L L 750mm
r = radius of gyration = = = 81.3 > 55
r c / 2 (18.44 mm )
I πc 4 4 c assumption was correct
= = 2
=
A πc 2
d = 2c = 36.9 mm
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13. • For L = 300 mm, assume L/r < 55
• Determine cylinder radius:
P   L 
σ all = = 212 − 1.585  MPa
A   r 
60 × 103 N   0.3 m  6
= 212 − 1.585  × 10 Pa
πc 2   c / 2 
c = 12.00 mm
• Check slenderness ratio assumption:
L L 300 mm
= = = 50 < 55
r c / 2 (12.00 mm )
assumption was correct
d = 2c = 24.0 mm
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14. Eccentric loading of columns
• Generally, columns are designed so
that the axial load is inline with the
column
• There are situations that the load will
be off center and cause a bending in
the column in addition to the
Pin-Pin Column
compression. This type of loading is
called eccentric load with Eccentric
Axial Load
• When a column is load off center,
bending can be sever problem and
may be more important than the
compression stress or buckling 14

15. Analysis of eccentric loads
• At the cut surface, there will be both an internal
moment, m, and the axial load P. This partial
section of the column must still be equilibrium,
and moments can be summed at the cut
surface, giving,
ΣM = 0
m + P (e + v) = 0
• bending in a structure can be modeled as m =
EI d2v/dx2, giving
EI d2v/dx2 + Pv = -Pe
• This is a classical differential equation that can
be solved using the general solution,
v = C2 sin kx + C1 cos kx - e
where k = (P/EI)0.5. The constants C1 and C2 can
be determined using the boundary conditions 15

16. • First, the deflection, v=0, at x = 0
0 = C2 0 + C1 1 - e
C1 = e
• The second boundary condition specifies the deflection, v=0, at X = L
0 = C2 sin kL + e cos kL - e
C2=e tan (kL/2)
• Maximum deflection
– The maximum deflection occurs at the column center, x = L/2, since both
ends are pinned.
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17. Maximum stress: secant formula
• Unlike basic column buckling, eccentric
loaded columns bend and must
withstand both bending stresses and
axial compression stresses.
• The axial load P, will produce a
compression stress P/A. Since the load
P is not at the center, it will cause a
bending stress My/I.
• The maximum moment, Mmax, is at
the mid-point of the column (x = L/2),
Mmax = P (e + vmax)
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18. • Combining the above equations gives
• But I = Ar2. This gives the final form of the secant formula as
• The stress maximum, σmax, is generally the yield stress or
allowable stress of the column material, which is known.
• The geometry of the column, length L, area A, radius of
gyration r, and maximum distance from the neutral axis c
are also known. The eccentricity, e, and material stiffness,
E, are considered known.
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19. 19

20. Design of columns under an eccentric load
• An eccentric load P can be replaced by a
centric load P and a couple M = Pe.
• Normal stresses can be found from
superposing the stresses due to the
centric load and couple,
σ = σ centric + σ bending
P Mc
σ max = +
A I
• Allowable stress method:
P Mc
+ ≤ σ all
A I
• Interaction method:
P A Mc I
+ ≤1
( σ all ) centric ( σ all ) bending
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21. Example
The uniform column consists of an 8-ft section
of structural tubing having the cross-section
shown.
a) Using Euler’s formula and a factor of safety
of two, determine the allowable centric load
for the column and the corresponding
normal stress.
b) Assuming that the allowable load, found in
part a, is applied at a point 0.75 in. from the
geometric axis of the column, determine the
horizontal deflection of the top of the
column and the maximum normal stress in
the column.
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22. SOLUTION:
• Maximum allowable centric load:
- Effective length,
Le = 2( 8 ft ) = 16 ft = 192 in.
- Critical load,
Pcr =
π 2 EI
=
( )(
π 2 29 × 106 psi 8.0 in 4 )
2
Le (192 in ) 2
= 62.1 kips
- Allowable load,
P 62.1 kips Pall = 31.1 kips
Pall = cr =
FS 2
P 31.1 kips
σ = all = σ = 8.79 ksi
A 3.54 in 2 22

23. • Eccentric load:
- End deflection,
 π P  
ym = e sec 
 2 P  − 1
  cr  
  π  
= ( 0.075 in ) sec  − 1
 2 2 
ym = 0.939 in.
- Maximum normal stress,
P  ec  π P 
σm = 1 + 2 sec
 2 P 

A r  cr  
31.1 kips  ( 0.75 in )( 2 in )  π 
= 2 
1+ sec 
3.54 in  (1.50 in ) 2  2 2 
σ m = 22.0 ksi
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24. Example
Determine the maximum flexural stress produced by a resisting Moment Mr of
+5000ft.lb if the beam has cross section shown in the figure.
Locate the neutral axis from the bottom end
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25. 25

26. • Work out the rest of example here
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