Determine the reactions and draw the shear and bending moment diagrams for the beam shown in Fig. 13.22(a) due to the loading shown and due to the support settlements of 40 mm at C and 25 mm at E. Use the method of consistent deformations.

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Accepted Answer

This beam was previously analyzed in Example 13.9 for the external loading by selecting the vertical reactions at the roller supports C and E as the redundants. We will use the same primary beam as used previously.

Support Settlements The specified support settlements are depicted in Fig. 13.22(b), from which it can be seen that the chord AE of the primary beam coincides with the undeformed position of the indeterminate beam; therefore, the settlements of supports C and E relative to the chord of the primary beam are equal to the prescribed settlements, that is

[latex]\Delta_{C R}=\Delta_{C}=-0.04 m[/latex] and [latex]\Delta_{E R}=\Delta_{E}=-0.025 m[/latex]

Compatibility Equations

[latex]\Delta_{C O}+f_{C C} C_{y}+f_{C E} E_{y}=\Delta_{C R}[/latex] (1)

[latex]\Delta_{E O}+f_{E C} C_{y}+f_{E E} E_{y}=\Delta_{E R}[/latex] (2)

Deflections of Primary Beam From Example 13.9,

[latex]\Delta_{C O}=-\frac{82,500 kN \cdot m ^{3}}{E I}=-\frac{82,500}{70\left(10^{6}\right(1,250)\left(10^{-6}\right)}=-0.943 m[/latex]

[latex]\Delta_{E O}=-\frac{230,000 kN \cdot m ^{3}}{E I}=-\frac{230,000}{70\left(10^{6}\right)(1,250)\left(10^{-6}\right)}=-2.629 m[/latex]

[latex]f_{C C}=\frac{333.333 m ^{3}}{E I}=\frac{333.333}{70\left(10^{6}\right)(1,250)\left(10^{-6}\right)}=0.00381 m / kN[/latex]

[latex]f_{E C}=f_{C E}=\frac{833.333 m ^{3}}{E I}=\frac{833.333}{70\left(10^{6}\right)(1,250)\left(10^{-6}\right)}=0.00952 m / kN[/latex]

[latex]f_{E E}=\frac{2,666.667 m ^{3}}{E I}=\frac{2,666.667}{70\left(10^{6}\right)(1,250)\left(10^{-6}\right)}=0.0305 m / kN[/latex]

Magnitudes of the Redundants By substituting the numerical values into the compatibility equations, we write

[latex]-0.943+0.00381 C_{y}+0.00952 E_{y}=-0.04[/latex] (1a)

[latex]-2.629+0.00952 C_{y}+0.0305 E_{y}=-0.025[/latex] (2a)

Solving Eqs. (1a) and (2a) simultaneously for [latex]C_y[/latex] and [latex]E_y[/latex], we obtain

[latex]C_{y}=107.6 kN \uparrow[/latex] and [latex]E_{y}=51.8 kN \uparrow[/latex]

Reactions and Shear and Bending Moment Diagrams The remaining reactions of the indeterminate beam can now be determined by equilibrium. The reactions and the shear and bending moment diagrams are shown in Fig. 13.22(c).

Question 13.15: Determine the reactions and draw the shear and bending momen...

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