Determine the reactions at the supports for the frame shown in Fig. 3.20(a).
Chapter 3
Q. 3.6
Step-by-Step
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
