Question

Determine the force in member DG of the loaded truss.

Answer 1

Concepts and reason

Truss:

A truss is a structure or frame that comprises two force members assembled in different configurations. The entire structure acts as a single member ultimately. A truss is an equilibrium body with the combination of various slender members to form a structured body. Method of joints:

It is a concept used to find the forces in the truss members according to which the sum of forces acting at a joint is equal to zero. Method of sections:

The method of sections is used to calculate the forces in members. It involves the cutting of the truss into sections and then solving the section for static equilibrium conditions. The section is cut such that the members sectioned are the members whose force is exposed on solving. A maximum of three forces (unknowns) can be solved using the three equilibrium equations from the sections' method. First, use moments and equilibrium conditions to form equations to calculate reaction force at the supports. Then use the section's method to cut the truss into sections and solve the section for moment condition. Use the joints' method to calculate the force in each member and assume the forces in that joint as tension, then use equilibrium conditions for forces and moments to form equations to calculate the force in each member. The type of force is decided based on the sign obtained in the final answer. If the answer has a negative sign, it is said to be in compression instead; if it has a positive sign, it is in tension.

Fundamentals

Support reactions:

If there is a force acting by support, the translational movement of the object is restricted or prevented. Then the force developed is called a support reaction. Main types of supports:

Fixed support:

It prevents rotational, horizontal, and vertical translation. It is shown in Figure \((1\) a) Pinned support:

This support prevents horizontal and vertical translation. It is shown in Figure \((1 \mathrm{~b})\). Roller support:

This support can't prevent the rotation or horizontal movement but can only provide vertical support reaction to the surface as shown in Figure Rocker support:

This support is almost similar to the rocker and will have an only vertical force acting as shown in Figure \((1 \mathrm{~d})\) Force moment:

The magnitude of the moment can be determined by the product of force and the perpendicular distance to the force. Write the equilibrium of forces along the \(x\) -axis. \(\sum F_{x}=0\)

Here, the sum of forces along the \(x\) -direction is \(\sum F_{x}\)

Write the equilibrium of forces along the \(\mathrm{y}\) -axis. \(\sum F_{y}=0\)

Here, the sum of forces along the y-direction is \(\sum F_{y}\). The moment is the product of force and its distance from the point. \(M=F d\)

Here, moment is \(M\), force is \(F\) and distance is \(d\). In equilibrium condition, the summation of moment acting about any point will be equal to zero. \(\sum M=0\)

General sign conventions for a moment:

The moment is considered as negative in a clockwise direction and positive in a counter-clockwise direction.

Draw the free body diagram as shown in Figure (2).

From Figure (2) calculate the moment about \(B\).

\(\sum M_{B}=0\)

\(-L(4 \mathrm{ft})-L(8 \mathrm{ft})-L(12 \mathrm{ft})-L(16 \mathrm{ft})-L(20 \mathrm{ft})+A_{y}(20 \mathrm{ft})=0\)

\(-L(60 \mathrm{ft})=-A_{y}(20 \mathrm{ft})\)

\(A_{y}=3 L\)

From Figure

(2) calculate the sum of vertical forces to obtain the reaction force \(B_{y}\)

\(\sum F_{y}=0\)

\(A_{y}+B_{y}=6 L\)

Substitute \(3 L\) for \(A_{y}\)

\(3 L+B_{y}=6 L\)

\(B_{y}=3 L\)

The free-body diagram is drawn to find the support reactions. The support reaction is determined by taking a moment about

point \(B\) and by the sum of forces along the vertical direction is calculated to obtain the vertical component of the reaction force at support B.

The free-body diagram of the left portion of the section \(a-a\) on the truss is shown in Figure (3).

From Figure (3), calculate the moment about point \(D\).

\(\sum M_{D}=0\)

\(-F_{F G}(3 \mathrm{ft})-L(4 \mathrm{ft})-L(8 \mathrm{ft})+A_{y}(8 \mathrm{ft})=0\)

Draw the forces acting at joint G as in Figure (4)

From Figure (4) apply equilibrium of horizontal forces acting on joint G.

\(\sum F_{x}=0\)

\(-F_{F G}+F_{G H} \cos \theta=0\)

\(F_{F G}=F_{G H} \cos \theta\)

From Figure \((4),\) obtain \((\theta)\) at the joint \(G\)

\(\tan \theta=\frac{1}{4}\)

\(\theta=\tan ^{-1}\left(\frac{1}{4}\right)\)

\(\theta=14.04^{\circ}\)

Substitute \(F_{G H} \cos \theta\) for \(F_{F G}\) in Equation (1) \(-\left(F_{G H} \cos \theta\right)(3 \mathrm{ft})-L(4 \mathrm{ft})-L(8 \mathrm{ft})+A_{y}(8 \mathrm{ft})=0\)

Plug-in \(14.04^{\circ}\) for \(\theta\), and \(3 L\) for \(A_{y}\) in Equation (2) \(-F_{G H}\left(\cos 14.04^{\circ}\right)(3 \mathrm{ft})-L(4 \mathrm{ft})-L(8 \mathrm{ft})+(3 L)(8 \mathrm{ft})=0\)

\(-F_{G H}(2.91 \mathrm{ft})+L(12 \mathrm{ft})=0\)

\(F_{G H}=4.12 L\)

The truss is sectioned on the left side. Properties of the right-angled triangle are used to calculate angle \(\theta\) and the member force.

\(F_{G I I}\) is calculated by taking a moment about point \(D\).

Write the equilibrium of forces in \(y\) -axis using Figure (4) . \(\sum F_{y}=0\)

\(-F_{G H} \sin \theta+F_{D G}=0\)

Substitute \(14.04^{\circ}\) for \(\theta\), and \(4.12 L\) for \(F_{G I I}\)

\(-(4.12 L)\left(\sin 14.04^{\circ}\right)+F_{D G}=0\)

\(F_{D G}=1 L\) (Tension)

The force in the \(D G\) member of loaded truss \(\left(F_{D G}\right)\) is \(1 L\) (Tension).

The force relation is obtained by taking the equilibrium of force along the vertical direction using Figure (4). Calculate the member force \(F_{D G}\) by substituting the angle \(\theta\) and force \(F_{G I I}\) in the relation.

The force in the \(D G\) member of loaded truss \(\left(F_{D G}\right)\) is \(1 L\) (Tension).