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Chapter 1- Static engineering systems1.1 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 beams1.2.2 standard section tables for rolled steel beams1.2.3 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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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 (y) and shear stresses xy(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,

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• 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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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 handbooks

              

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Euler’s Formula for Pin-Ended Beams

Putting

v vv

vl

l

v v

6

7

8

9

Design of columns under centric loads

• Experimental data demonstrate

- for large Le/r, cr follows Euler’s formula and depends upon E but not Y.

- for intermediate Le/r, cr depends on both Y and E.

- for small Le/r, cr is determined by the yield strength Y and not E.

(le/k)2

le/k

le/k

le/k

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Structural Steel

American Inst. of Steel Construction

• For Le/r > Cc

92.1

/ 2

2

FS

FSrL

E crall

ecr

• For Le/r > Cc

3

2

2

/

8

1/

8

3

3

5

2

/1

c

e

c

e

crall

c

eYcr

C

rL

C

rLFS

FSC

rL

• At Le/r = Cc

YcYcr

EC

22

21 2

le/k

le/k

le/k

le/k

le/k le/k

le/k le/k

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Sample problem

Using the aluminum alloy2014-T6, 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

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 necessary.

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2

4

gyration of radius

radiuscylinder

2

4 c

c

c

A

I

r

c

• For L = 750 mm, assume L/r > 55

• Determine cylinder radius:

mm44.18

c/2

m 0.750

MPa 103721060

rL

MPa 10372

2

3

2

3

2

3

cc

N

A

Pall

• Check slenderness ratio assumption:

553.81mm 18.44

mm750

2/

c

L

r

L

assumption was correct

mm 9.362 cd

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• For L = 300 mm, assume L/r < 55

• Check slenderness ratio assumption:

5550mm 12.00

mm 003

2/

c

L

r

L

assumption was correct

mm 0.242 cd

• Determine cylinder radius:

mm00.12

Pa102/

m 3.0585.1212

1060

MPa 585.1212

62

3

c

cc

N

r

L

A

Pall

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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 compression. This type of loading is called eccentric load

• When a column is load off center, bending can be sever problem and may be more important than the compression stress or buckling

Pin-Pin Column

with Eccentric Axial Load

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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 - ewhere k = (P/EI)0.5. The constants C1 and C2 can be determined using the boundary conditions

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• 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 - eC2=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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• 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)

Maximum stress: secant formula

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• 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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20

• Allowable stress method:

allI

Mc

A

P

• Interaction method:

1bendingallcentricall

IMcAP

• 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,

I

Mc

A

P

bendingcentric

max

Design of columns under an eccentric load

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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.

Example

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SOLUTION:

• Maximum allowable centric load:

- Effective length,

kips 1.62

in 192

in 0.8psi 10292

462

2

2

ecr

L

EIP

- Critical load,

2in 3.54

kips 1.31

2

kips 1.62

A

P

FS

PP

all

crall

kips 1.31allP

ksi 79.8

- Allowable load,

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• Eccentric load:

in. 939.0my

122

secin 075.0

12

sec

crm P

Pey

- End deflection,

22sec

in 1.50

in 2in 75.01

in 3.54

kips 31.1

2sec1

22

2

cr

m P

P

r

ec

A

P

ksi 0.22m

- Maximum normal stress,

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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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• Work out the rest of example here