Determine the force in each member of the truss shown in Fig. 4.20(a) by the method of joints.

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Static Determinacy The truss is composed of 7 members and 5 joints and is supported by 3 reactions. Thus, m + r = 2j. Since the reactions and the members of the truss are properly arranged, it is statically determinate.

From the dimensions of the truss given in Fig. 4.20(a), we find that all inclined members have slopes of 12:5. Since joint E has two unknown non-collinear forces, [latex]F_{CE}[/latex] and [latex]F_{DE}[/latex], acting on it, we can begin the method of joints without first calculating the support reactions.

Joint E Focusing our attention on joint E in Fig. 4.20(b), we observe that in order to satisfy [latex]\sum{F_x} = 0[/latex], the horizontal component of [latex] F_{DE} [/latex] must push to the left into the joint with a magnitude of 25 kN to balance the 25 kN external load acting to the right. The slope of member DE is 12:5 , so the magnitude of the vertical component of [latex]F_{DE}[/latex] is (12/5)(25), or 60 kN. Thus, the force in member DE is compressive, with a magnitude of

[latex]F_{DE} = \sqrt{(25)^2 + (60)^2} = 65 kN[/latex]

[latex]F_{DE} = 65 kN (C)[/latex]

With the vertical component of [latex]F_{DE}[/latex] now known, we can see from the figure that in order for [latex]\sum{F_y} = 0[/latex] to be satisfied, [latex]F_{CE}[/latex] must pull downward on joint E with a magnitude of 60 - 30 = 30 kN.

[latex]F_{CE} = 30 kN (T)[/latex]

Joint C Next, we consider the equilibrium of joint C. Applying [latex]\sum{F_x} = 0[/latex] , we obtain [latex]F_{CD}[/latex].

[latex]F_{CD} = 50 kN (C)[/latex]

From [latex]\sum{F_y} = 0[/latex], we obtain [latex]F_{AC}[/latex].

[latex]F_{AC} = 30 kN (T)[/latex]

Joint D Both of the unknown forces, [latex]F_{AD}[/latex] and [latex]F_{BD}[/latex], acting at this joint have inclined directions, so we draw the free-body diagram of this joint as shown in Fig. 4.20(c) and determine the unknowns by solving the equilibrium equations simultaneously:

[latex]+\longrightarrow \sum{F_x} = 0[/latex] [latex]50 + \frac{5}{13}(65) - \frac{5}{13} F_{AD} + \frac{5}{13} F_{BD} = 0[/latex]

[latex]+\uparrow \sum{F_y} = 0[/latex] [latex]-\frac{12}{13}(65) -\frac{12}{13}F_{AD} -\frac{12}{13}F_{BD} = 0[/latex]

Solving these equations simultaneously, we obtain

[latex]F_{AD} = 65 kN[/latex] and [latex]F_{BD} = -130 kN[/latex]

[latex]F_{AD} = 65 kN (T)[/latex]

[latex]F_{BD} = 130 kN (C)[/latex]

Joint B (See Fig. 4.20(b).) By considering the equilibrium of joint B in the horizontal direction ([latex]\sum{F_x}=0[/latex]) , we obtain [latex]F_{AB}[/latex].

[latex]F_{AB} = 50 kN (T)[/latex]

Having determined all the member forces, we apply the remaining equilibrium equation ([latex]\sum{F_y}=0[/latex]) at joint B to calculate the support reaction [latex]B_y[/latex].

[latex]B_y = 120 kN \uparrow[/latex]

Joint A By applying [latex]\sum{F_x}=0[/latex], , we obtain [latex]A_x[/latex].

[latex]A_x = 75 kN \longleftarrow [/latex]

From [latex]\sum{F_y}=0[/latex], we obtain [latex]A_y[/latex].

[latex]A_y = 90 kN \downarrow [/latex]

Checking Computations To check our computations, we consider the equilibrium of the entire truss. Applying the three equilibrium equations to the free body of the entire truss shown in Fig. 4.20(b), we obtain

[latex]+\longrightarrow \sum{F_x} = 25 + 50 - 75 = 0[/latex] Checks

[latex]+\uparrow \sum{F_y} = -30 - 90 + 120 = 0[/latex] Checks

[latex]+\circlearrowleft \sum{M_B} = 30(5) - 25(12) - 50(6) + 90(5) = 0[/latex] Checks

Question 4.5: Determine the force in each member of the truss shown in Fig...

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