Download - Engineering science lesson 5
1
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)
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 (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,
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
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 handbooks
5
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
10
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
11
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.
12
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
13
• 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
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 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
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 - ewhere k = (P/EI)0.5. The constants C1 and C2 can be determined using the boundary conditions
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 - eC2=e tan (kL/2)
• Maximum deflection– The maximum deflection occurs at the column center, x = L/2, since both
ends are pinned.
17
• 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
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.
19
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
21
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
22
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,
23
• 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,
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
25
26
• Work out the rest of example here