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Q. 3.6

Determine the reactions at the supports for the frame shown in $Fig. 3.20(a)$.

Verified Solution

Free-Body Diagram See Fig. 3.20(b).

Static Determinacy The frame is internally stable with $r = 3$. Therefore, it is statically determinate.

Support Reactions

$+\longrightarrow \sum{F_x}=0$

$(\frac{2+3}{2})(26)(\frac{5}{13})+ C_x=0$

$C_x = -25 k$

$C_x= 25 k \longleftarrow$                            Ans.

$+\circlearrowleft \sum{M_A}=0$

$-2(26)(13) -\frac{1}{2}(1)(26)(\frac{26}{3}) – 50(24 + 12) + 25(10) + C_y(48) = 0$

$C_y = 48.72 k$

$C_y = 48.72 k \uparrow$                   Ans.

$+\uparrow \sum{F_y}=0$

$A_y – (\frac{2+3}{2})(26)(\frac{12}{13}) – 50 + 48.72 = 0$

$A_y = 61.28 k$

$A_y = 61.28 k \uparrow$                        Ans.

Checking Computations

$+\circlearrowleft \sum{M_B} = -61.28(24) + 2(26)(13) +\frac{1}{2}(1)(26)(\frac{2}{3})(26) – 50(12) + 48.72(24)$

$= -0.107 k-ft \approx 0$                          Checks