Calculate the end moments at the supports in the beam shown in Fig. 16.42 if the support at B is subjected to a settlement of 12 mm. Furthermore, the second moment of area of the cross section of the beam is 9 × 10^6 mm^4 in the span AB and 12 × 10^6 mm^4 in the span BC; Young’s modulus, E, is

Question

Calculate the end moments at the supports in the beam shown in Fig. 16.42 if the support at B is subjected to a settlement of [latex]12 \mathrm{~mm}[/latex]. Furthermore, the second moment of area of the cross section of the beam is [latex]9 imes 10^{6} \mathrm{~mm}^{4}[/latex] in the span [latex]\mathrm{AB}[/latex] and [latex]12 imes 10^{6} \mathrm{~mm}^{4}[/latex] in the span [latex]\mathrm{BC}[/latex]; Young's modulus, [latex]E[/latex], is [latex]200000 \mathrm{~N} / \mathrm{mm}^{2}[/latex].

Accepted Answer

In this example the FEMs produced by the applied loads are modified by additional moments produced by the sinking support. Thus, using Table 16.6

[latex] \begin{aligned} & M_{\mathrm{AB}}^{\mathrm{F}}=-\frac{6 imes 5^{2}}{12}-\frac{6 imes 200000 imes 9 imes 10^{6} imes 12}{\left(5 imes 10^{3} ight)^{2} imes 10^{6}}=-17.7 \mathrm{kN} \mathrm{m} \ & M_{\mathrm{BA}}^{\mathrm{F}}=+\frac{6 imes 5^{2}}{12}-\frac{6 imes 200000 imes 9 imes 10^{6} imes 12}{\left(5 imes 10^{3} ight)^{2} imes 10^{6}}=+7.3 \mathrm{kN} \mathrm{m} \end{aligned} [/latex]

Since the support at [latex]\mathrm{C}[/latex] is an outside pinned support, the effect on the FEMs in BC of the settlement of B is reduced (see the last case in Table 16.6). Thus

[latex] \begin{aligned} & M_{\mathrm{BC}}^{\mathrm{F}}=-\frac{40 imes 6}{8}+\frac{3 imes 200000 imes 12 imes 10^{6} imes 12}{\left(6 imes 10^{3} ight)^{2} imes 10^{6}}=-27.6 \mathrm{kN} \mathrm{m} \ & M_{\mathrm{CB}}^{\mathrm{F}}=+\frac{40 imes 6}{8}=+30.0 \mathrm{kN} \mathrm{m} \end{aligned} [/latex]

The DFs are

[latex]\mathrm{DF}_{\mathrm{BA}}=\frac{K_{\mathrm{BA}}}{K_{\mathrm{BA}}+K_{\mathrm{BC}}}=\frac{\left(4 E imes 9 imes 10^{6} ight) / 5}{\left(4 E imes 9 imes 10^{6} ight) / 5+\left(3 E imes 12 imes 10^{6} ight) / 6}=0.55[/latex]

Hence

[latex]\mathrm{DF}_{\mathrm{BC}}=1-0.55=0.45[/latex]

[latex] \begin{array}{lcccc} \hline & A & &B &&C \ \hline { DFs } & – & 0.55 && 0.45 & 1.0\ \hline ext { FEMs } & -17.7 & +7.3 && -27.6 & +30.0\ ext { Balance C }&&&&& _{\swarrow} \underline{-30.0} \ ext{Carrv over} & & && -15.0 & \ ext{Balance B} & & _{\swarrow} \underline{+19.41} && \underline{+15.89} & \ ext{Carry over} & +9.71 & & & \ ext{Final moments} & -7.99 & +26.71& & -26.71 & 0 \ \hline \end{array}[/latex]

Note that in this example balancing the beam at [latex]\mathrm{B}[/latex] has a significant effect on the fixing moment at [latex]\mathrm{A}[/latex]; we therefore complete the distribution after a carry over to A.

	FEMs
Load case	$\textstyle{M}_{\mathrm{AB}}^{\mathrm{F}}$	$\textstyle{M}_{\mathrm{BA}}^{\mathrm{F}}$
	$-{\frac{W L}{\mathrm{8}}}$	$+{\frac{W L}{\mathrm{8}}}$
	$-{\frac{W a b^{2}}{L^2}}$	$+{\frac{W a^{2} b}{L^2}}$
	$-{\frac{w L^{2}}{12}}$	$+{\frac{w L^{2}}{12}}$
	$-\frac{w}{L^{2}}\left[\frac{L^{2}}{2}(b^{2}-a^{2})-\frac{2}{3}L(b^{3}-a^{3})+\frac{1}{4}(b^{4}-a^{4})\right]$	$+\frac{w b^{3}}{L^{2}}\left(\frac{L}{3}-\frac{b}{4}\right)$
	$+\frac{M_{0}b}{L^{2}}(2a-b)$	$+\frac{M_{0}a}{L^{2}}(2b-a)$
	$-{\frac{6EIδ}{L^2}}$	$-{\frac{6EIδ}{L^2}}$
	0	$-{\frac{3EIδ}{L^2}}$

Question 16.20: Calculate the end moments at the supports in the beam shown ...

Related Answered Questions

Calculate the forces in the members of the truss shown in Fig.16.19(a). All members have the same cross sectional area, A, and Young’s modulus, E. ...

Verified Answer:

Calculate the end moments in the members of the frame shown in Fig. 16.49. All members have the same flexural rigidity, EI; note that the member CD is pinned to the foundation at D. ...

Verified Answer:

Obtain the bending moment diagram for the portal frame shown in Fig. 16.47(a). The flexural rigidity of the horizontal member BC is 2EI while that of the vertical members AB and CD is EI. ...

Verified Answer:

Obtain the bending moment diagram for the frame shown in Fig. 16.44; the flexural rigidity EI is the same for all members. ...

Verified Answer:

Calculate the support reactions in the beam shown in Fig. 16.40; the flexural rigidity, EI, of the beam is constant throughout. ...

Verified Answer:

Calculate the reactions at the supports in the beam ABCD shown in Fig. 16.41. The flexural rigidity of the beam is constant throughout. ...

Verified Answer:

Determine the end moments in the members of the portal frame shown in Fig. 16.36; the second moment of area of the vertical members is 2.5I while that of the horizontal members is I. ...

Verified Answer:

Find the support reactions in the three-span continuous beam shown in Fig. 16.33. ...

Verified Answer:

Determine the support reactions in the parabolic arch shown in Fig. 16.29 assuming that the second moment of area of the cross section of the arch varies in accordance with the secant assumption. ...

Verified Answer:

Determine the support reactions in the semicircular two-pinned arch shown in Fig. 16.27(a). The flexural rigidity, EI, of the arch is constant throughout. ...

Verified Answer: