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Transcript of 7 Combined Loadings
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DMV 4343JAN ~ JUN `07
DEPARTMENT MANUFACTURING / PRODUCT DESIGN /MOULD / TOOL AND DIE
SEMESTER 4
COURSE MECHANICS OF MATERIALS DURATION 8 hrs
COURSE CODE DMV 4343 / DMV 5343 REF. NO.
VTOS NAME MISS AFZAN BINTI ROZALIMR RIDHWAN BIN RAMELI
PAGE 13
TOPICCOMBINED LOADINGS
SUB TOPIC7.1 Stress Elements for Axially Loaded Bar, Shaft in Torsion, and Beam in Bending7.2 Combined Loadings and Complex Stresses
Chapter 7 COMBINED LOADINGS p1
INFORMATION SHEET
REF NO. :PAGE :
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Chapter 7 COMBINED LOADINGS p2
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7.1 Stress Elements for Axially Loaded Bar, Shaft in Torsion, and Beam in Bending
In previous chapters we developed methods for determining the stress distributions in a
member subjected to either an internal axial force, a shear force, a bending moment, or a
torsional moment. Most often, however, the cross section of a member is subjected to
several of these types of loadings simultaneously, and as a result, the method of
superposition, if it applies, can be used to determine the resultantstress distribution caused
by the loads. For application, the stress distribution due to each loading is first determined,
and then these distributions are superimposed to determine the resultant stress distribution.
The principle of superposition can be used for this purpose provided a linear relationship
exists between the stress and the loads. Also, the geometry of the member should not
undergo significant change when the loads are applied. This is necessary in order to ensure
that the stress produced by one load is not related to the stress produced by any other load.
The discussion will be confined to meet these two criteria.
FIGURE 7.1 This chimney is subjected
to the combined loading of wind and
weight. It is important in investigate the
tensile stress in the chimney since
masonry it weak in tension
PROCEDURE FOR ANALYSIS
The following procedure provides a general means for establishing the normal and shear
stress components at a point in a member when the member is subjected to several different
types of loadings simultaneously. It is assumed that the material is homogeneous and
behaves in a linear-elastic manner. Also, Saint-Venant's principles require that the point
where the stress is to be determined is far removed from any discontinuities in the cross
section or points of applied load.
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Internal Resultants.
Section the member perpendicular to its axis at the point where the stress is to be
determined and obtain the resultant internal normal and shear force components and
the bending and torsional moment components.
The force components should act through the centroidof the cross section, and the
moment components should be computed about centroidal axes, which represent the
principal axes of inertia for the cross section.
This,of course, involves drawing free-body diagrams and writing equilibrium
equations.
Average Normal Stress
1. Compute the stress component associated with eachinternal loading. For each case,represent the effect either as a distribution of stress acting over the entire cross-
sectional area, or show the stress on an element of the material located at a
specified point on the cross section.
TABLE 7.1 FORMULA FOR STRESSES
FORCE, LOADINGS FORMULA
NORMAL FORCEThe internal normal force is developed by a uniform
normal-stress distribution
= P I A.
SHEAR FORCEThe internal shear force in a member that is subjected to
bending is developed by a shear-stress distribution
= VQ/It
BENDING
MOMENT
For straight members the internal bending moment is
developed by a normal-stress distribution that varies
linearly from zero at the neutral axis to a maximum at the
outer boundary of the member
= - My/I
TORSIONAL
MOMENT
For circular shafts and tubes the internal torsional
moment is developed by a shear-stress distribution that
varies linearly from the central axis of the shaft to a
maximum at the shaft's outer boundary
= Tp/J
If the member is a closed thin-walled tube = T/2Amt
THIN-WALLED
PRESSURE
VESSELS
If the vessel is a thin-walled cylinder, the internal
pressure p will cause a biaxial state of stress in the
material
2 = pr/2t
If the vessel is a thin-walled sphere, then the biaxial state
of stress is represented by two equivalent components
2 = pr/2t
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EXAMPLE 7.1
An axial compressive load of 800 kN is applied eccentrically to a short rectangular
compression member, as shown below.
An eccentrically loaded short compression member.
Determine
(a) Individual normal stress for the axially loaded member, and
(b) Sketch the stress distribution of each individual normal stress.
Plan the Solution
The eccentric load Pproduces axial deformation plus bending
about the y and z axes. Therefore, this problem involves asuperposition of the stresses due to F, My, and Mz.Solution
Stress Resultants: Figure 2 shows the stress resultants on
cross section ABCD. Applying equilibrium to a free-body
diagram of the member above section ABCD we get the
following expressions for the stress resultants:
F = - PMy = - P dz
Mz = - P dy
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Individual Normal Stresses: Combining all the above equations for the normal stress due to
the axial force F, for the normal stress due to the bending-moment components, we have:
x =F
+Myz
-Mzy
A Iy Iz
Substitute individual stress into the previous resultant
stress equation, we obtain the following:
(x)F =
- P
=
- 800 kN
= -1000 PaA
(40m)
(20m)
The stress contributions as in Figure (a) (a)
(x)My =(-Pdz)z
=(-800 kN)(5m)z
= -150 z kN/m3Iy (1/12)(40m)(20m)
3
Maximum stress occurs at the farthest point along y-axis
that is at z = 10;
Hence
(x)My = -150 z kN/m3 = 1500 kN/m3
The stress contributions as in Figure (b) (b)
(x)Mz =(-Pdy)y
=(-800 kN)(10m)z
= -75 y kN/m3Iy (1/12)(20m)(40m)
3
Maximum stress occurs at the farthest point along z-axis
that is at y = 20;
Hence
(x)My = -75 y kN/m3 = 1500 kN/m3
The stress contributions as in Figure (c)
(c)
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EXAMPLE 7.2 (Continuation of Example 7.1)
Refer Example 7.1.
Determine
(a) The distribution of normal stress on a cross section, say ABCD, that is far enough
from the point of load application that concentration effects may be neglected.
(b) Sketch the stress distribution, and
(c) Identify the location of the neutral axis in cross sectionABCD
Superposition of Stresses:
(a) Using Figure (a) through (c) from solution of Example 7.1, we can combine, algebraically,
the individual stress contributions at four corners to get
(x)A = - 1000 + 1500 - 1500 = - 1000 N/m2
(x)B = - 1000 - 1500 - 1500 = - 4000 N/m2
(x)C = - 1000 - 1500 + 1500 = - 1000 N/m2
(x)D = - 1000 + 1500 + 1500 = + 2000 N/m2
(b)
+ +
(a) (b) (c)
=
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(c) The location of point R and S can be determined from
x (y,z) = - 1000 - 75y - 150z = 0
To find R (we know y = -20)x (y,z) = - 1000 - 75(-20) - 150z = 0
z =-500
= 3.33-150
To find S (we know z = -10)
x (y,z) = - 1000 - 75y- 150(-
10)= 0
y =-500
= 6.67-75
So;
R (-20m, 3.33m)
S (6.67m, -10m)
Chapter 7 COMBINED LOADINGS p9
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7.2.2 Combined Axial Loading and Torsion.
The next example problem illustrates the superposition of stresses due to combined axial
and torsional loading. The solution involves application of Mohrs circle. Following is the
procedure for constructing and using Mohrs circle of stress
Draw Mohr's Circle
1. Establish a set of (, ) axes, with the same scale on both axes. Remember, the
+ axis points downward. It is good idea to use paper that has a grid, like graph
paper or '"engineering paper." Use a scale that will result in a circle of reasonable
size.
2. Assuming that x, y, and xy are given (or can be determined from a given stress
element), locate point X at (x, xy) and point Y at (y, -xy),
3. Connect points X and Y with a straight line, and locate the center of the circle
where this line crosses the axis at (avg, 0).
4. Draw a circle with center at (avg, 0) and passing through points X and Y. It is best
to use a compass to draw the circle.
Compute the required information
5. Form the triangle with sides xy and (x - y)/2, and compute
R = (x - y)2
+ xy226. If the stresses on a particular face, call it face n, are required, locate point N on
the circle by turning an angle 2 counterclockwise (or clockwise) on the circle,
corresponding to rotating an angle counterclockwise (clockwise) from some
reference face on the stress element. Using trigonometry, calculate n and nt
7. If the principal stresses and the orientation of the principal planes are required,
use to calculate the principal stresses, and use trigonometry to determine some
angle, such as 2xp1 that can be used to locate a principal plane, say p 1, with
respect to some known face, say the x face.
8. Use a procedure similar to Step 7 if the maximum in-plane shear stress and the
planes of maximum shear stress are required.
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In this section we will discussed an example which combined more than two types of
loadings. Following example combined normal stress, bending stress and torsional stress on
a member.
EXAMPLE 7.4
Wind blowing on a sign produces a pressure whose resultant,
P, acts in the -y direction at point C, as shown. The weight of
the sign, Ws, acts vertically through point C, and the thin-wall
pipe that supports the sign has a weight Wp.
Determine the principal stresses at points A and B, where the
pipe column is attached to its base. Use the following
numerical data.
Pipe OD = 3.5 m, A = 2.23 m2, Iy = Iz = 3.02m4, Ip = 6.03m
4
Ws = 125 N, Wp = 160 N, P = 75 N, b = 40 m L = 220 m
Plan the Solution
It will be a good idea to tabulate the stress resultants, stress
formulas, and so forth, so that no stress contribution will be
missed. The weight Wscontributes to the axial force, and it also
produces a moment about the y axis. The wind force P
produces a transverse shear force in the ydirection, and it alsocauses a torque about the xaxis and a moment about the z
axis. A correct free-body diagram is essential.
Figure (a)
Solution
Stress Resultants: All six stress resultants on the cross section at the base of the pipe are
shown in Figure (a). The upper portion of Figure (a) can serve as a free-body diagram for
determining these six stress resultants. Let us tabulate the equilibrium equations and indicate
what stress is produced by each stress resultant and label each individual stress.
Individual Stresses: Using the formulas from Table 7.1, we can compute the numerical value of
each of the nonzero stresses listed in Table 1.
A1 = B1 =F
=
- (125N)
(160N) = -128 Pa (1)
A 2.23 m2
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In summary, the principal stresses at points A and B, rounded to three significant figures are:
(1)A =9510
Pa; (2)A = 0 ; (3)A = - 80 Pa
(1)B = 267 Pa ; (2)B = 0 ; (3)B = -3290 Pa
Review the Solution
By showing all six possible internal resultants at the cross section where stresses are to be
calculated, by writing down and solving all six possible equilibrium equations, and by
carefully considering what stress(es) is (are) produced by each stress resultant, we have
accounted for the effects of all loads on the structure. As noted earlier, we have been careful
to make sure that each stress component acts in the direction that "makes sense." For
example, the force Pbends the pipe in the direction that produces tension at pointA, and soforth.
The maximum flexural stress at the base occurs at neitherA norB.
Chapter 7 COMBINED LOADINGS p15