Tag Archives: equilibrium of truss members

Statics 3.6 – Equilibrium of Truss Members Connected to a Gusset Plate | Hibbeler 14th Edition

The gusset plate is subjected to the forces of three members. Determine the tension force in member C and its angle θ for equilibrium. The forces are concurrent at point O. Take F=8 kN.

Engineering Mechanics: Statics 14th Edition by RC Hibbeler Problem 3-5 Equilibrium of Truss Members Connected in a Gusset Plate
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Statics 3.2 – Equilibrium of Truss Members that are Pin Connected | Hibbeler 14th Edition

The members of a truss are pin connected at joint O. Determine the magnitude of F1 and its angle θ for equilibrium. Set F2=6 kN.

Figure 3.1: Engineering Mechanics: Statics Equilibrium of Particle
Figure 3.1/3.2

Solution:

Free-body diagram:

Free-body-diagram-for-Problem-3.2 of Engineering Mechanics: Statics by Russell C. Hibbeler

Equations of Equilibrium:

The summation of forces in the x-direction:

∑Fx=06sin⁡70°+F1cos⁡θ−5cos⁡30°−45(7)=0F1cos⁡θ=4.2920(1)\begin{aligned} \sum F_x & = 0 &\\ 6 \sin 70 \degree + F_1 \cos \theta - 5 \cos 30 \degree - \dfrac {4}{5} \left(7 \right) & = 0 & \\ F_1 \cos \theta & = 4.2920 & (1) \end{aligned}

The summation of forces in the y-direction:

∑Fy=06cos⁡70°+5sin⁡30°−F1sin⁡θ−35(7)=0F1sin⁡θ=0.3521(2)\begin{aligned} \sum F_y & =0 & \\ 6 \cos 70 \degree+5 \sin 30 \degree - F_1 \sin \theta - \dfrac{3}{5} \left( 7 \right ) & =0 & \\ F_1 \sin \theta &=0.3521 & (2)\\ \end{aligned}

We came up with 2 equations with unknowns F1 F_1 and θ \theta . To solve the equations simultaneously, we can use the method of substitution.

Using equation 1, solve for F1 F_1 in terms of θ \theta .

F1cos⁡θ=4.2920F1=4.2920cos⁡θ(3)\begin{aligned} F_1 \cos \theta & = 4.2920 &\\ F_1 & =\dfrac{4.2920}{ \cos \theta } & (3) \\ \end{aligned}

Now, substitute this equation (3) to equation (2).

F1sin⁡θ=0.3521(4.2920cos⁡θ)sin⁡θ=0.35214.2920⋅sin⁡θcos⁡θ=0.35214.2920tan⁡θ=0.3521tan⁡θ=0.35214.2920θ=tan⁡−10.35214.2920θ=4.69°\begin{aligned} F_1 \sin \theta & = 0.3521 \\ \left ( \dfrac {4.2920}{\cos \theta} \right) \sin \theta & =0.3521 \\ 4.2920 \cdot \dfrac{\sin \theta}{\cos \theta} & = 0.3521 \\ 4.2920 \tan \theta & = 0.3521 \\ \tan \theta & = \dfrac{0.3521}{4.2920} \\ \theta &= \tan ^{-1} \dfrac{0.3521}{4.2920} \\ \theta & = 4.69 \degree \end{aligned}

Substitute the solved value of θ \theta to equation (3).

F1=4.2920cos⁡θF1=4.2920cos⁡4.69°F1=4.31kN\begin{aligned} F_1 & = \dfrac{4.2920}{\cos \theta} \\ F_1 &= \dfrac{4.2920}{\cos 4.69 \degree} \\ F_1 & = 4.31 \text{kN} \end{aligned}

Therefore, the answers to the questions are:

F1= 4.31 kNθ= 4.69°\begin{aligned} F_1= & \:4.31 \: \text {kN} \\ \theta = & \: 4.69 \degree \end{aligned}

Statics 3.1 – Equilibrium of Truss Members in Pin Connection | Hibbeler 14th Edition

The members of a truss are pin connected at joint O. Determine the magnitudes of F1 and F2 for equilibrium. Set θ=60.

Figure 3.1: Engineering Mechanics: Statics Equilibrium of Particle
Figure 3.1

Solution:

Free-body diagram:

Free-Body Diagram for Problem 3.1 of Engineering Mechanics: Statics 14th Edition by Russell C. Hibbeler

Equations of Equilibrium:

Take the sum of horizontal forces considering forces to the right positive, and equate to zero.

∑Fx=0F1cos⁡60°+F2sin⁡70°−5cos⁡30°−45(7)=00.5F1+0.9397F2=9.9301(1)\begin {aligned} \sum{F}_x &= 0 & \\ F_1 \cos{60 \degree}+F_2 \sin{70 \degree}-5\cos{30 \degree}-\dfrac{4}{5}\left(7\right) &= 0 &\\ 0.5F_1+0.9397F_2&=9.9301 &(1)\\ \end {aligned}

Take the sum of vertical forces considering upward forces positive, and equate to zero.

∑Fy=0−F1sin⁡60°+F2cos⁡70°+5sin⁡30°−35(7)=0−0.8660F1+0.3420F2=1.7(2)\begin{aligned} \sum F_y&=0 &\\ -F_1\sin60\degree+F_2\cos70\degree+5\sin30\degree-\dfrac{3}{5}\left(7\right)&=0 &\\ -0.8660F_1+0.3420F_2&=1.7 &(2)\\ \end{aligned}

Now, we have two equations with two unknowns F1 F_1 and F2 F_2 . So, we have a system of two equations. We can solve this using algebra, or we can directly use our calculator with this capability. The answers are

F1=1.83 kNF2=9.60 kNF_1=1.83 \: \text{kN}\\ F_2=9.60 \: \text{kN}