A beam is simply supported over a length of L = 100 mm and is loaded at its center with a transverse load of P = 50 N. Its cross section has width b = 15 mm and depth 2c = 10 mm. The material has linear viscoelastic behavior with elastic and steady-state creep strains, similar to the model of

Question

A beam is simply supported over a length of L = 100 mm and is loaded at its center with a transverse load of P = 50 N. Its cross section has width b = 15 mm and depth 2c = 10 mm. The material has linear viscoelastic behavior with elastic and steady-state creep strains, similar to the model of Fig. 15.27(a), with the constants being [latex]E_{1} = 3[/latex] GPa and [latex]η_{1} = 10^5[/latex] GPa·s. Determine (a) the maximum stress in the beam, (b) the initial elastic deflection when the load is applied, and (c) the deflection after one week.

Accepted Answer

Due to the linear viscoelastic behavior, stresses may be obtained from the ordinary elastic bending formula:

[latex]\sigma=\frac{M c}{I}, \quad I=\frac{2 b c^3}{3}[/latex]

Here, I is from Fig. A.2(a) in Appendix A. The maximum stress is at the midlength, where the moment is M = PL/4 from Fig. A.4(a). Hence, substituting for M and I gives

[latex]\sigma=\frac{3 P L}{8 b c^2}=\frac{3(50 N )(100 mm )}{8(15 mm )(5 mm )^2}=5.00 MPa[/latex]

Using the equation from Fig. A.4(a), we find that the initial elastic deflection is controlled by [latex]E_{1}[/latex], as the creep strain is initially zero:

[latex]v=\frac{P L^3}{48 E_1 I}=\frac{(50 N )(100 mm )^3}{48(3000 MPa )\left(1250 mm ^3\right)}=0.278 mm[/latex]

To obtain the deflection as affected by creep, note that the strain–time behavior is given by the equation from Fig. 15.27(a):

[latex]\varepsilon=\frac{\sigma}{E_1}+\frac{\sigma t}{\eta_1}[/latex]

Hence, the time-dependent modulus for t = 1 week = 604,800 seconds is

[latex]E(t)=\frac{\sigma}{\varepsilon}=\frac{1}{\frac{1}{E_1}+\frac{t}{\eta_1}}=\frac{1}{\frac{1}{3000 MPa }+\frac{604,800 s }{10^8 MPa \cdot s }}=156.7 MPa[/latex]

Repeating the deflection calculation with this lower value E(t) finally gives the deflection after one week:

[latex]v=\frac{P L^3}{48 E(t) I}=\frac{(50 N )(100 mm )^3}{48(156.7 MPa )\left(1250 mm ^3\right)}=5.32 mm[/latex]

Question 15.7: A beam is simply supported over a length of L = 100 mm and i...

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