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2003-01-18 - Division of Solid Mechanics

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Examination in Damage <strong>Mechanics</strong> and Life Analysis (TMHL61)<br />

LiTH <strong>2003</strong>-<strong>01</strong>-<strong>18</strong><br />

Part 1<br />

3. (1 point) Constant-amplitude fatigue strengths for materials<br />

subjected to different cyclic loading conditions are<br />

<strong>of</strong>ten expressed in Haigh diagrams. Consider the five<br />

a<br />

loading cases where the loading varies sinusoidally<br />

A<br />

and<br />

1. the mean value σ m is constant,<br />

200<br />

100<br />

B 2. the amplitude σ a is constant,<br />

3. the maximum value σ max (= σ m + σ a ) is constant,<br />

0<br />

0<br />

m<br />

100 200 300 400 MPa<br />

4. the minimum value σ min (= σ m − σ a ) is constant,<br />

and<br />

5. the stress ratio R = σ min /σ max is constant.<br />

In the Haigh diagram given, two straight lines (A<br />

and B) are shown. For each line one <strong>of</strong> the five<br />

loading conditions applies.<br />

(a) Determine which one <strong>of</strong> the five loading variables is constant for line A shown in<br />

the figure (i.e., match line A to one <strong>of</strong> the five loading conditions). Explain your choice.<br />

(b) Determine which one <strong>of</strong> the five loading variables is constant for line B shown in<br />

the figure (i.e., match line B to one <strong>of</strong> the five loading conditions). Explain your choice.<br />

Solution and answer here:<br />

For line A one notices that σ m + σ a is a constant (= 300 MPa). Thus, line A gives that<br />

stress σ max is constant.<br />

For line B one notices that σ m − σ a = σ min is constant (= 200 MPa). Thus, stress σ min is<br />

constant.<br />

Answer:<br />

(a) Line A: stress σ max is constant, (b) line B: stress σ min is constant.<br />

4. (1 point)<br />

To determine the fatigue limit <strong>of</strong> a material, the so-called stair-case method is<br />

sometimes used. Explain the principles, requirements and assumptions <strong>of</strong> this method<br />

(no formulas are needed).<br />

Solution and answer here:<br />

The fatigue testing is performed so that if a test specimen fails (due to fatigue), the next<br />

specimen will be tested at a lower stress level. If it does not fail (becomes a run-out),<br />

the next specimen will be tested at a higher stress lever. Then, assuming that the yield<br />

limit has a normal distribution, and knowing the number <strong>of</strong> failures and run-outs at each<br />

stress level, the mean value and the standard deviation <strong>of</strong> the fatigue limit can be<br />

estimated. This requires that tests performed at a stress level too far away from the<br />

mean value should be rejected.<br />

7 <strong>2003</strong>-<strong>01</strong>-23/TD

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