Truss MCQ Quiz - Objective Question with Answer for Truss - Download Free PDF

Concept:

In a pin-jointed plane frame in equilibrium, if three forces act at a joint and are in equilibrium, Lami’s Theorem is applicable:

\( \frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma} \)

Given:

Joint C is under equilibrium with three forces:
• Force in member AC
• Force in member BC
• A vertical downward load of 6 kN

Coordinates for triangle:
• AB = 8 m
• AC = BC = 42+32=5" id="MathJax-Element-13-Frame" role="presentation" style="position: relative;" tabindex="0">42+32−−−−−−√=542+32=5" id="MathJax-Element-1-Frame" role="presentation" style="position: relative;" tabindex="0">42+32−−−−−−√=542+32=5 42+32=5 $\sqrt{4^2 + 3^2} = 5$ m
So, angle θ = tan⁻¹(3/4)

At joint C, all three forces meet:

AC and BC are inclined at θ = tan⁻¹(3/4) = 36.87°

Angle between AC and BC = 180° – 2θ = 106.26°

Angle between AC and vertical load = 90° + θ = 126.87°

Angle between BC and vertical load = 90° + θ = 126.87°

Using Lami's Theorem at Joint C:

\( \frac{T_{AC}}{\sin(126.87°)} = \frac{T_{BC}}{\sin(126.87°)} = \frac{6}{\sin(106.26°)} \)

Using:

• sin(126.87°) = sin(53.13°) ≈ 0.799
• sin(106.26°) ≈ 0.961

Now, calculate the force in member AC:

\( T_{AC} = \frac{6 \times \sin(126.87°)}{\sin(106.26°)} = \frac{6 \times 0.799}{0.961} = 4.99 \approx 5 \, \text{kN} \)

Force in member AC = 5 kN (Tensile)

Explanation:

Method of Sections vs. Method of Joints

In truss analysis, the Method of Sections offers a key advantage over the Method of Joints by allowing the direct calculation of forces in any member without the need to analyze the entire truss. This can be particularly useful in large or complex trusses where analyzing each joint would be time-consuming and cumbersome. Here, we analyze the given options:

1. "The Method of Sections does not require the assumption of concurrent forces at the joints." (Incorrect)

   • Both the Method of Sections and the Method of Joints assume that the forces at the joints are concurrent and follow the principles of static equilibrium.

2. "The Method of Sections can directly calculate the forces in any member without having to analyze the entire truss." (Correct)

   • This method allows you to "cut" through the truss and analyze a section, which simplifies the process of finding forces in specific members.

3. "The Method of Sections is more accurate than the Method of Joints." (Incorrect)

   • Both methods are equally accurate as they are based on the same principles of static equilibrium.

4. "The Method of Sections is the only method that allows for the calculation of reaction forces." (Incorrect)

   • Reaction forces can also be calculated using the Method of Joints or by applying the equilibrium equations to the entire truss.

Explanation:

At joint C

Equilibrium in x-direction

∑Fx = 0

F_CD = 0

Equilibrium in y-direction

∑Fy = 0

F_CB - 10 = 0

FCB = 10 (tensile)

Concept:

Perfect truss:

A truss that has got enough members to resist the loads without undergoing deformation in its shape is called a perfect truss. The triangular truss is the simplest perfect truss and it has three joints and three members.

For a perfect truss m = 2j - 3

Where,

the total number of members = m

total number of joints = j

Deficient Truss:

In a deficient truss, the number of members in it is less than that required for a perfect truss. Such trusses cannot retain their shape when loaded.

Redundant trusses:

In a redundant truss, the number of members in it is more than that required in a perfect truss. Such trusses cannot be analyzed by making use of the equations of equilibrium alone. Thus, a redundant truss is statically indeterminate.

Explanation:

• A space truss is a three-dimensional truss structure where members are connected in multiple planes to distribute loads efficiently.

• Transmission towers are a classic example of space trusses as they consist of interconnected steel members forming a three-dimensional framework to support high-voltage power lines.

Additional Information

1. Rafters  → These are typically two-dimensional elements used in roof structures.

2. Bicycle frame  → While it has a structural framework, it does not follow a truss system.

3. Roofing  → Most roofing structures use planar trusses, which are two-dimensional.

Concept:

For a truss, Degree of static indeterminacy = m + r - 2j

Where,

m = number of members, r = number of reactions, and j = number of Joints

Calculation:

In the given truss,

Number of members(m) = 11, 

Number of Joints(j) = 6,

number of reactions(r) = 4

Degree of static indeterminacy = m + r - 2j

= 11 + 4 - (2 × 6)

= 15 - 12

= 3.

Hence, In the figure, the static degree of indeterminacy is 3.

Concept:

Rule for zero member force:

• At a joint, if three members are meeting, two members are collinear (same line), then force in the third member is always zero (if there is no external load at that joint)
• At a joint, if two non-collinear members are meeting with no external load at that joint then forces in both members will be zero.

Calculation:

(i) At joint C, we get member CB to have zero force member as the load is acting in the horizontal direction and no load is in the direction of CB.

(ii) Joints A and B do not move relative to each other i.e. member AB has no change length

⇒ No strain ⇒ zero Force

Concept:

Let us use the method of joints 

Consider Joint S

∑F_y = 0

F_RS - P = 0

F_RS = P

Since F_RS is acting away from joint S it is tensile.

Consider Joint R

∑Fy = 0

F_TR sin45^° - F_RS = 0

⇒ F_TR = P × √ 2

Since F_TR is acting towards the joint R it is compressive

∑F_x = 0

F_TR cos 45^° - F_QR = 0

F_QR = P

Since F_QR is acting away from the joint it is tensile.

Concept:

The force in the member AB will be determined by method of joints, i.e. by analyzing each joints.

For this we will be fixing some rule, they are

• for every joint we will consider the unknown forces are going away from the joint
• and the  forces are considered positive.
• So after determination, if one forces comes positive then, it will mean that it is tensile and if it comes negative then it would mean that it is compressive.

Calculations:

Taking moment about A, we get - 

R_C × 6 = 5 × 6 sin 60

⇒ R_C = 5 sin 60

⇒  RC = \(\frac{{5\sqrt 3 }}{2}\)

Analyzing joint C:

R_C + R_CB sin 60 = 0

⇒ RCB sin 60 = - RC 

⇒ RCB × \(\frac{{\sqrt 3 }}{2}\) = - \(\frac{{5\sqrt 3 }}{2}\)

⇒ RCB = - 5 kN [-ve denotes that the force will be toward the joint, hence it will be compressive in nature]

Explanation:

By using the method of the section, cut the section which is perpendicular to AB and FE. 

Taking moment about E =0

F_AB × 4 - W × 3 - W ×  9 = 0

Hence F_AB = 3W . 

Explanation:

Truss: It is defined as a framework, typically consisting of rafters, posts, and struts, supporting a roof, bridge, or other structure.

Truss can be classified as follows:

a) Simple Truss: It consists of a series of triangles so that the weight being supported is distributed evenly to the supports.

∴ A truss will be classified as a simple truss if m = 2j – 3.

b) Complex Truss: It is not necessarily a set of the triangle as the members may overlap each other.

c) Compound Truss:

It is formed by connecting 2 or more simple truss together. Often, this type of truss is used to support loads acting over a larger span as it is cheaper to construct a lighter compound truss than a heavier simple truss.

Various type of compound truss are as follows:

Type 1: The trusses may be connected by a common joint & bar.

Type 2: The trusses may be joined by 3 parallel bars.

Type 3: The trusses may be joined where bars of a large simple truss, called the main truss, have been substituted by simple truss, called secondary trusses.

For the given truss:

B (m = number of member)  = 5, R (r = number of reactions) = 3 and J (j = number of joint) = 4

Static indeterminacy (D_s) = External static indeterminacy (D_se) + Internal static indeterminacy (Dsi)

Dse = R - 3 = 3 - 3 = 0

Dsi = m + 3 - 2 j = 5 + 3 - 2 × 4 = 0

Ds = Dse + Dsi = 0 + 0 = 0

∴ Then given truss is statically determinate.

Confusion Points

The stability of the member cannot be confirmed as sometimes the determinate structure may be unstable. e.g. a beam with three linear roller support.

The portal usually is a sway frame extending between a pair of trusses whose purpose also is to transfer the reactions from a lateral-bracing truss to the end posts of the trusses, and, thus, to the foundation.

It provides additional stability to the structure and sometimes it may also be meant to resist traverse load along with stability.

Concept:

To identify the zero force members in truss, the following rules are required to be used:

1. If at a joint three members are meeting, two of which are collinear then force in third member is zero provide that no external load is acting on joint.

2. If two non-collinear members are meeting at joint and no external load is acting on that joint, then force in both members is zero.

Calculation:

Joint D:

Member DC and DE are non-collinear and No external load an joint D.

So

\({F_{DE}} = {F_{DC}} = 0\)  Rule 2

Joint E:

∑F_y = 0

⇒ F_EF = 0

∑ F_x = 0

Joint C:

F_E2 = + 5 kN (Tensile)

F_CI = 0  Rule (2)

F_CB = 0 Rule (2)

Zero force members

Area DE, DC, EF, CI, CB

Tube are shown in above diagram.

Concept:

The strain energy stored in truss,

U = U_OA + U_OB

\(U = \frac{{P_1^2L}}{{2AE}} + \frac{{P_2^2L}}{{2AE}}\)

The deflection at O will be, \(\Delta = \frac{{\partial U}}{{\partial P}}\)

Calculation:

At joint O

ΣF_x = 0, F_OA = F_OB..............(i)

ΣF_y = 0, F_OA × cosθ + F_OB × cosθ = P ..........(ii)

2F_OA × cosθ = P

Here, cosθ = 3/4.24 and P = 10 kN

\({F_{OA}} = \frac{P}{{2\cos \theta }} = \frac{P}{{2 \times \frac{3}{{4.24}}}} = 0.7067P\)

The strain energy stored in truss,

U = UOA + U_OB

\(U = \frac{{F_{OA}^2L}}{{2AE}} + \frac{{F_{OB}^2L}}{{2AE}} = \frac{{2F_{OA}^2L}}{{2AE}} = \frac{{F_{OA}^2L}}{{AE}}\)

⇒ \(U = \frac{{{{(0.7067P)}^2}L}}{{AE}} = \frac{{0.5{P^2}L}}{{AE}}\)

The deflection at O will be, \(\Delta = \frac{{\partial U}}{{\partial P}}\)

\(\Delta = \frac{{0.5 \times 2PL}}{{AE}} = \frac{{PL}}{{AE}}\)

\(\Delta = \frac{{10 \times {{10}^3} \times 4.24 \times {{10}^3}}}{{100 \times 2 \times {{10}^5}}} = 2.12\ mm\)