4a osna obremenitev primer
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4a-ex. Axial Member ExamplesEx. 4a.1Ex. 4a.2Ex. 4a.3
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Example 4a.1
Given: A stepped-bar (areas A1, A2) made of two
materials (moduli E1, E2), acted on by forces 2W(upward at Pt. B) and W (downward at Pt. E).
Req'd: Determine the elongation of the bar Dtotal.
Sol'n:
Step 1. Equilibrium. Solve for internal forces between
points where P(x), A(x) or E(x) change.
PAB = W; PBC = PCD = PDE = W Bar broken up into lengths withconstant force, area and modulus.
Step 2. Elongation/Force Relationship. To solve for the elongation, break the bar into lengths
(segments) over which all the values (force, area and modulus) are constant.
The top such segment, AB, has length a, area A1, modulus E1, and internal force PAB = W
(compression). The change in length of AB is:
Since AB is in compression, it shortens: Da
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Example 4a.2
Given: A truncated conical pillar supports a
weight W.
Req'd: Determine the shortening of the column
due to the load. Neglect the weight of the
column. Measure x from the top of the pillar.
Sol'n: Break up the column into lengths where
the P, A and E are all constant.
Here, it is differential length dx. Over dx, the
force, area and modulus are essentially
constant. By finding the elongation dD of each
dx,
and adding them up - integrating - we find the
deflection of the bar.
P(x) = W (compression) and E(x)=E are
constant over L.
Area A(x) varies from pR2 at the top to p(4R)2 at
the bottom. We must find an equation for
A(x)=p[R(x)]2, which is done at the right.
The length with constant force, area and modulus is
dx. The stress at any point is s(x) = P(x)/A(x).
Radius as a function of x:
At the endpoints:
R(x=0) = R;
R(L) = 4R.
R increases linearly from R to 4R:
R(x) = R[1+3x/L].
Knowing A(x), sum all of the dD over length L:
The elongation is negative - the column shortens. The effective stiffness for the truncated cone
acted on by constant force throughout its length is Keff= |W/D| = (4pR2E)/L
Example 4a.3
Given: A stepped bar constrained on both ends is
subjected to force P at Point C. Due to the force, Point C
moves to the right.
Req'd: Use the displacement method to determine (a)
the force P, and (b) the stiffness of the bar K, in terms of
the displacement d, Young's modulus E, and the
geometric parameters of the bar (area and length).
x. Examples http://strengthandstiffness.com/4_basic/page_4a-ex.ht
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Sol'n: Solve this problem using the three steps spelled out for the displacement method.
Step 1. Compatibility. The geometric constraint of the system requires that the total change in
length of the system be zero:
DAC + DCB = 0
or:
DAC = DCB = d
Step 2. Hooke's Law. The forces in the two members can be related to the displacements as
follows:
Step 3. Equilibrium. Equilibrium requires that: P + FAC = FCB
Combining the three equations gives:
The stiffness of this system is:
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Updated: 05/18/2009 DJD
x. Examples http://strengthandstiffness.com/4_basic/page_4a-ex.ht
d 3 27 7 2013 22:48