Lecture - 04 & 05

Analysis of Trusses: Simple, Compound, and Space Truss

Trusses

A truss is an assembly of straight members connected at their ends by flexible joints to form a rigid structure.

ü Key Features:

· Lightweight yet strong.

· Members usually made of steel, aluminum, or wood.

· Connections to gusset plates are typically bolted or welded.

ü Common Uses:

· Support for bridges and building roofs.

· Structural frameworks in space stations.

Plane Trusses:

All members and applied loads lie in a single plane.
Commonly used for bridge decks and building roofs.
Example:

Roof trusses connected by purlins, which transfer roof loads (self-weight, snow, wind) to truss joints.

Can be treated as two-dimensional structures for analysis.

Space Trusses:

Members are arranged in three dimensions.
Cannot be treated as plane trusses due to shape, arrangement, or loading.
Examples: Transmission towers, latticed domes etc..

Some of the common configurations of bridge and roof trusses, many of which have been named after their original designers

Common Bridge Trusses:

Common Roof Trusses:

Assumptions for Analysis of Trusses

To simplify analysis, the following assumptions are made:

 Frictionless Connections

All members are connected only at their ends.
Plane trusses use frictionless hinges.
Space trusses use frictionless ball-and-socket joints.

 Joint-Only Loading

All loads and support reactions are applied only at the joints.

 Straight Centroidal Axis

The centroidal axis of each member coincides with the line joining the centers of adjacent joints.

 Small Displacements

All displacements are small relative to the lengths of the truss members, allowing linear analysis.

Purpose of Assumptions in Truss Analysis and Ideal vs. Real Truss Behavior

To simplify the analysis of truss structures, several idealizing assumptions are adopted. These assumptions lead to a theoretical model known as an ideal truss, where each member is subjected solely to axial forces (tension or compression), without any bending moment or shear force.

Justification of Ideal Truss Assumptions

An ideal truss is based on the following key assumptions:

1. All members are connected only at their ends by frictionless hinges (in plane trusses) or ball-and-socket joints (in space trusses).

2. External loads and support reactions are applied only at the joints, never along the length of any member.

3. The centroidal axis of each member aligns precisely with the straight line connecting the centers of the adjacent joints.

4. Displacements are small relative to the member lengths, ensuring that the geometry does not significantly change under load.

These assumptions imply that each truss member functions as a two-force member. it experiences only two equal and opposite forces applied at its ends. Since the forces are equal in magnitude, opposite in direction, and collinear, the member remains in static equilibrium under purely axial loading.

This can be shown through the equilibrium equations:

∑Fx = 0: No net force in the horizontal direction.
∑Fy = 0: No net force in the vertical direction.
∑M = 0: No net moment about any point.

Because of this equilibrium, the member resists either:

Axial tension (when the member elongates), or
Axial compression (when the member shortens).

These internal axial forces, determined under the idealized assumptions, are called primary forces.

Real Trusses and Secondary Forces

In actual engineering practice, the idealized behavior of trusses is rarely fully achieved due to practical constraints and construction methods. Some of the primary deviations include:

Member Connections: Real trusses use welded or bolted gusset plates instead of frictionless hinges. This introduces partial restraint at joints.
Continuous Members: Some members may pass through joints and act as continuous beams.
Distributed Loads: Although loads from floor beams or purlins are typically transferred to joints, the self-weight of the truss members acts along their length, introducing bending and shear.
Connection Geometry: Misalignment in member centroids at joints can introduce unintended moments.

These deviations from the ideal assumptions result in secondary forces, which include:

Bending moments
Shear forces
Additional axial effects due to connection eccentricity or continuity

These secondary forces complicate the internal stress distribution within the members.

Design Consideration of Secondary Forces

Although secondary forces are inherently present in real trusses, they can often be minimized and are generally neglected in preliminary design because:

Truss members are typically slender, reducing the magnitude of bending moments.
Joint design can be optimized so that member centroidal axes are concurrent at a point, minimizing moment transfer.

In most practical applications, the primary forces dominate the structural behavior of the truss, and thus the structure can be designed effectively using the ideal truss model.

Arrangement of Members of Plane Trusses—Internal Stability

Internal vs. External Stability of a Plane Truss

A plane truss is considered internally stable if the number and geometric arrangement of its members are such that, when detached from its supports, the truss retains its shape and behaves as a rigid body without undergoing deformation.

The term "internal" refers specifically to the arrangement and connectivity of the members within the truss.
Instability caused by an insufficient number of supports or by an improper arrangement of supports is classified as external instability.

Simple Trusses

· The simplest truss unit consists of a triangular arrangement of three members.

· A simple truss is formed by enlarging the basic truss element, which contains three members and three joints

· so the total number of members m in a simple truss is given by

m = 2j -3

j = total number of joints (including those attached to the supports).

Compound Trusses

A compound truss is created by joining two or more simple trusses into a single rigid structure.
Combining trusses allows for longer spans, greater load capacity, or complex geometries that cannot be achieved efficiently with a single simple truss.
To ensure the compound truss behaves as a single rigid body (no relative movement between component trusses), the connection between the simple trusses must be able to transmit at least three independent force components.
These force components:

Must not be all parallel.
Must not be concurrent.

Internal Stability Criterion for Plane Trusses

the minimum number of members required for a simple plane truss:

m=2j−3

where:

· m = total number of members

· j = total number of joints (including those connected to supports)

General stability condition for any plane truss:

If a truss has m members and j joints:

· m < 2j – 3, the truss is internally unstable

· m ≥ 2j – 3, the truss is internally stable

· The condition is necessary but not sufficient for internal stability.

· Even if the member count satisfies the equation, the geometric arrangement must also ensure rigidity.

Examples

Classify each of the plane trusses shown in Figures below as internally stable or unstable.

Equations of Condition for Plane Trusses

Three types of connection arrangements commonly used to connect two rigid trusses to form a single (internally unstable) truss.

In Fig. below, two rigid trusses, AB and BC, are connected together by an internal hinge at B. Because an internal hinge cannot transmit moment, it provides an equation of condition:

Another type of connection arrangement is shown in Fig. below. This involves connecting two rigid trusses, AB and CD, by two parallel members. Since these parallel (horizontal) bars cannot transmit force in the direction perpendicular to them, this type of connection provides an equation of condition:

A third type of connection arrangement involves connecting two rigid trusses, AB and CD, by a single link, BC, as shown in Fig. below. Since a link can neither transmit moment nor force in the direction perpendicular to it, it provides two equations of condition:

Static Determinacy, Indeterminacy, and Instability of Plane Trusses

The conditions of static instability, determinacy, and indeterminacy of plane trusses can be summarized as follows:

· m = number of members

· j = number of joints

· r = number of external reactions

· Statically indeterminate trusses have more members andyor external reactions than the minimum required for stability. The excess members and reactions are called redundants, and the number of excess members and reactions is referred to as the degree of static indeterminacy, i, which can be expressed as

· i = (m+ r)-2 j

Examples

Classify each of the plane trusses shown in Fig. below as unstable, statically determinate, or statically indeterminate. If the truss is statically indeterminate, then determine the degree of static indeterminacy.

Identification of Zero-Force Members in Trusses

In truss analysis, it is common to encounter members that carry zero force under a specific loading condition. This occurs because trusses are typically designed to withstand multiple loading scenarios, and certain members may be unloaded in a particular case. Zero-force members are also intentionally incorporated to:

Provide bracing to compression members against buckling.
Prevent vibration in slender tension members.

Identifying zero-force members by inspection can significantly reduce the computational effort in truss analysis. Two common configurations resulting in zero-force members are:

1. Two Noncollinear Members at a Joint without External Load or Reaction

o If only two noncollinear members meet at a joint with no external load or support reaction, both members carry zero force.

o Reason: From equilibrium (ΣFy=0 and ΣFx=0), both members must have zero axial force.

2. Three Members at a Joint, Two Being Collinear, without External Load or Reaction

o If three members meet at a joint, with two of them being collinear and no external load or reaction applied, the noncollinear member carries zero force.

o Reason: The noncollinear member cannot have a vertical (or perpendicular) force component without an external force to balance it, thus its axial force is zero.

Note: These rules apply only when there is no external load or reaction at the joint.

Procedure for Analysis – Method of Joints

1. Check Static Determinacy & Stability

o Verify if the truss is statically determinate and stable.

o If not, stop the analysis (indeterminate trusses require different methods).

2. Identify Zero-Force Members (By Inspection)

o Look for joints with:

§ Two non-collinear members and no external load/reaction → both are zero-force members.

§ Three members, where two are collinear and no load/reaction is applied → the third member is a zero-force member.

3. Determine Slopes of Inclined Members

o Calculate the slopes/angles of inclined members (except zero-force members) for use in equilibrium equations.

4. Draw the Free-Body Diagram (FBD) of the Whole Truss

o Show all external loads and support reactions.

o Mark the zero-force members with a “0”.

5. Choose the starting joint

· Examine the FBD and look for a joint with no more than two unknown member forces.

o The two unknowns must not be collinear so that the x- and y-equilibrium equations are independent.

· If such a joint exists, start solving there using:

∑Fx=0 and ∑Fy=0

· If no joint has ≤2 unknowns, first determine the support reactions for the entire truss by applying:

∑Fx=0, ∑Fy=0, ∑M=0

· After reactions are known, re-inspect joints to find one with ≤2 unknowns and proceed.

· Sometimes, starting at a joint with three unknowns is possible if:

o One member is a zero-force member by inspection, or

o Two unknowns are collinear and one component is already known.

6. Analyze the Selected Joint

o Draw the FBD of that joint.

o Assume all unknown forces are in tension (arrows pointing away from the joint).

o Apply joint equilibrium equations:

§ ΣFx = 0

§ ΣFy = 0

o Interpret results:

§ Positive value → member is in tension.

§ Negative value → member is in compression.

o Shortcut: If one unknown is purely horizontal or vertical, solve quickly by inspection.

7. Move to the Next Joint

o Choose another joint with ≤ 2 unknown forces.

o Repeat Step 6 until all forces are found.

8. Check Your Work

o If reactions were found in Step 5, verify them using unused joint equilibrium equations.

o If reactions were found from joint analysis, check them using the overall truss equilibrium equations.

o All checks must satisfy equilibrium for a correct analysis.

Method of Joints– Warren Truss

Example

1. Determine the force in each member of the Warren truss shown in Fig. by the method of joints.

Given / Geometry

• Warren truss, panel length = 20 ft, height = 15 ft; four panels (total span 80 ft).

• Joint loads: 24 k at B, 30 k at C, and 12 k at D (acting downward).

• Supports: pin at A, roller at E.

• All diagonals form 3–4–5 triangles ⇒ sinθ = 3/5, cosθ = 4/5.

Step 1: Support Reactions

Take counterclockwise moments as positive.

· ∑M_A = 0: 80·E_y – 24X20 – 30X40 – 12X60 = 0 ⇒ E_y = 30 k

· ∑F_y = 0: A_y + E_y − (24 + 30 + 12) = 0 ⇒ A_y = 36 k

· ∑F_x = 0: A_x = 0

Step 2: Joint A

Unknowns: AB, AF. Assume tension positive (members pull away from the joint).

· ∑F_y = 0: 36 + (3/5) X F_AF = 0 ⇒ F_AF = −60 k (compression)

· ∑F_x = 0: F_AB + (4/5)·F_AF = 0 ⇒ F_AB = 48 k (tension)

Step 3: Joint B

Unknowns: BC, BF (AB known).

· ∑F_x = 0: −48 + F_BC = 0 ⇒ F_BC = 48 k (tension)

· ∑F_y = 0: −24 + F_BF = 0 ⇒ F_BF = 24 k (tension)

Step 4: Joint F

Unknowns: FG, FC (AF and BF known). For F_AF = −60 k (compression), the force on joint F is (+48 i + 36 j) k.

· ∑F_y = 0: 36 − 24 − (3/5)·F_FC = 0 ⇒ F_FC = 20 k (tension)

· ∑F_x = 0: 48 + F_FG + (4/5)·F_FC = 0 ⇒ F_FG = −64 k (compression)

Step 5: Joint G

Unknowns: GH, GC (FG known).

· ∑F_x = 0: (+64) + F_GH = 0 ⇒ F_GH = −64 k (compression)

· ∑F_y = 0: −F_GC = 0 ⇒ F_GC = 0 k (zero‑force)

Step 6: Joint E

Unknowns: DE, HE (E_y known).

· ∑F_y = 0: 30 + (3/5)·F_HE = 0 ⇒ F_HE = −50 k (compression)

· ∑F_x = 0: −F_DE − (4/5)·F_HE = 0 ⇒ F_DE = 40 k (tension)

Step 7: Joint D

Unknowns: CD, DH (DE known; external 12 k downward).

· ∑F_x = 0: +40 − F_CD = 0 ⇒ F_CD = 40 k (tension)

· ∑F_y = 0: −12 + F_DH = 0 ⇒ F_DH = 12 k (tension)

Step 8: Joint H

Unknowns: CH (GH, HE, and DH known).

· ∑F_y = 0: −12 + 30 − (3/5)·F_CH = 0 ⇒ F_CH = 30 k (tension)

· Check ∑F_x = 0: (+64) − 40 − (4/5)·30 = 0 ✓

Step 9: Joint C (Equilibrium Check)

Forces meeting at C: BC (48 T), CD (40 T), FC (20 T), CH (30 T), GC (0), load 30 k ↓.

· ∑F_x = 0: (−48) + 40 + (−16) + 24 = 0 ✓

· ∑F_y = 0: (+12) + (+18) − 30 = 0 ✓

Step 10: Final Member Forces

Support Reactions

A_x = 0, A_y = 36 k, E_y = 30 k

Example

2. Determine the force in each member of the truss shown in Fig. below by the method of joints.

Given truss with external loads:

• 30 kN downward at E
• 25 kN to the right at E
• 50 kN to the right at C

Support at A (pinned) and B (roller). Geometry based on 5–12–13 triangles.

Static Determinacy

m = 7, j = 5, r = 3

m + r = 2j

1) Support Reactions

∑Fx = 0: Ax - 25 - 50 = 0 → Ax = -75 kN (75 kN to the left)
∑MA = 0: 5By - (50)(6) - (25)(12) = 0 → By = 120 kN (up)
∑Fy = 0: Ay + By - 30 = 0 → Ay = -90 kN (90 kN down)

2) Method of Joints

Joint B

∑Fy: 120 + F_BD (12/13) = 0 → F_BD = -130 kN (Compression)
∑Fx: -F_BA + F_BD (5/13) = 0 → F_BA = 50 kN (Tension)

Joint E

∑Fx: 25 + F_ED (5/13) = 0 → F_ED = -65 kN (Compression)
∑Fy: -30 - F_EC - F_ED (12/13) = 0 → F_EC = 30 kN (Tension)

Joint C

∑Fx: 50 + F_CD = 0 → F_CD = -50 kN (Compression)
∑Fy: -F_CA + F_CE = 0 → F_CA = 30 kN (Tension)

Joint D

From equilibrium:
F_DA = 65 kN (Tension)
F_DC = -50 kN (Compression)

3) Member Forces

4) Support Reactions

Ax = -75 kN (left)
Ay = -90 kN (down)
By = 120 kN (up)

Method of Sections – Summary of Principles

Objective

The method of sections is employed to determine the internal forces in selected members of a truss directly.
This approach eliminates the need to compute numerous intermediate member forces, as is often required in the method of joints.

Fundamental Concept

An imaginary section is introduced through the truss, intersecting the members in which the internal forces are to be determined.
This section divides the truss into two distinct parts.

Analytical Constraints

Since only three independent equilibrium equations are available, the method can generally determine a maximum of three unknown member forces per section.
Consequently, the section should normally be chosen to intersect no more than three members with unknown forces.

Special Considerations

In certain truss configurations, it may be possible to employ a section intersecting more than three unknown members if the geometry or loading conditions allow direct determination of one or more of these forces.
Such cases typically arise in symmetrical trusses or when specific members carry zero force under the given loading conditions.

Procedure for Analysis of Statically Determinate Plane Trusses by the Method of Sections

1. Select the Section

o Choose a section that passes through the members whose forces you need to find.

o Do not cut through more than three members with unknown forces.

o The section should divide the truss into two separate parts.

2. Choose the Simpler Portion for Analysis

o You can use either side of the cut for calculations.

o Pick the side that requires less computation.

o To avoid calculating support reactions:

§ If one portion has no support reactions, choose that side.

§ If both portions have supports, first calculate reactions using equilibrium equations for the entire truss, then choose the side with fewer external loads and reactions.

3. Draw the Free-Body Diagram (FBD)

o Show all external loads and reactions acting on the chosen portion.

o Indicate the cut member forces as tensile (arrows pulling away from joints).

o Clearly label all forces and dimensions.

4. Apply Equilibrium Equations to Find Unknowns

o Use the three standard equations of equilibrium:

∑Fx=0, ∑Fy=0, ∑M=0

o To avoid solving simultaneous equations:

§ Arrange calculations so each equation has only one unknown.

§ Consider using alternative equilibrium equation sets (e.g., moments about different points) to simplify the process.

5. Check Your Results

o Use an alternative equilibrium equation (different from the ones used to solve forces) for verification.

o This equation should ideally involve all three calculated member forces.

o If correct, the equation will be satisfied within acceptable tolerance.

Alternative Forms of Equations of Equilibrium of Plane Structures

Although the usual equilibrium equations provide the most convenient means of analyzing a majority of plane structures, the analysis of some structures can be expedited by employing one of the following two alternative forms of the equations of equilibrium:

in which A and B are any two points in the plane of the structure, provided that the line connecting A and B is not perpendicular to the q axis, and

in which A, B, and C are any points in the plane of the structure, provided that these three points do not lie on the same straight line.

Example:

1. Determine the forces in members CD, DG, and GH of the truss shown in Fig. below by the method of sections.

Solution

1. Selecting the Section

A section aa is drawn cutting through three members of interest: CD, DG, and GH.
This cut divides the truss into two portions:

Left portion: ACGE
Right portion: DHI

To avoid calculating support reactions, we choose the right-hand portion (DHI) for analysis.

2. Drawing the Free-Body Diagram (FBD)

The FBD of portion DHI is prepared.
All three unknown member forces (F_CD, F_DG, and F_GH) are:

Assumed tensile (arrows pulling away from joints).
Clearly marked on the diagram.

The slope of the inclined force F_DG is indicated for use in calculations.

The negative answer for F_CD indicates that our initial assumption about this force being tensile was incorrect, and F_CD is actually a compressive force.

F _ CD = 80 k (C)

Checking Computations

Example

2. Determine the forces in members CJ and IJ of the truss shown in Fig. below by the method of sections.

Solution

1. Selecting the Section

A section aa is drawn passing through members IJ, CJ, and CD.
This cut divides the truss into two portions:

Left portion: ACI
Right portion: DGJ

The left-hand portion (ACI) is chosen for analysis.

2. Determining Support Reactions

Before calculating member forces, find the support reactions using the equilibrium of the entire truss.
From Fig. (b):

Ax=0,

Ay=50 k (upward),

Gy=50 k (upward)

3. Drawing the Free-Body Diagram (FBD)

Prepare the FBD for portion ACI of the truss.
Show:

All external forces (including support reactions at A).
The three unknown member forces (F_IJ, F_CJ, F_CD) as tensile (arrows pulling away from joints).
The slopes of the inclined forces F_IJ and F_CJ, determined from truss dimensions in Fig.(a).

4. Applying Equilibrium Equations

Since F_CJ and F_CD pass through point C, taking moments about point C eliminates them from the equation, leaving only F_IJ:

The negative answer for F_IJ indicates that our initial assumption about this force being tensile was incorrect. Force F_IJ is actually a compressive force.

F_IJ = 65.97 k (C)

Now,

Pick the moment center: Take moments about point O, the intersection of the lines of action of F_JF and F_CD

This choice eliminates both F_JF and F_CD from the moment equation (their moment arms about O are zero).

Geometry for the moment arm

The slope of member IJ is 1:4

The distance OC = 4(IC) =4(25) = 100 ft

Equilibrium of moments about O yields

Checking Computations. To check our computations, we apply an alternative equation of equilibrium, which involves the two member forces just determined.

Determine the forces in members FJ, HJ, and HK of the K truss shown in Fig.(a) by the method of sections.

Method of Sections for the K-Truss

 A horizontal section aa through FJ, HJ, HK also cuts FI → that’s 4 unknown member forces.

 You only have 3 equilibrium equations ⇒ aa alone is insufficient. (1 section is not sufficient)

Use a curved section bb around the middle joint

Take a curved cut bb (as in Fig.(a)) that slices FI, IJ, JK, HK and choose the upper portion I–K–N–L as the free body (to avoid support reactions).
Key idea: pick moment centers where (nearly) all other cut members’ lines of action pass through, so they drop out of the moment sum.

first compute FHK by considering section bb and then use section aa to determine F FJ and FHJ

From section bb to find F_HK:

Take moments about point I (the intersection of the lines of action of the other cut members on this FBD).

Section aa. The free-body diagram of the portion IKNL of the truss above section aa is shown in Fig.(c). To determine F_HJ,

sum moments about F, which is the point of intersection of the lines of action of F_FI and F_FJ. Thus,

By summing forces in the horizontal direction, we obtain

Checking Computations. Finally, to check our calculations, we apply an alternative equilibrium equation, which involves the three member forces determined by the analysis. Using Fig.(c), we write

Analysis of Compound Trusses Using Both Methods

Two main methods:

Method of Joints – Solves forces in truss members by isolating individual joints.
Method of Sections – Solves forces by cutting through the truss and analyzing a section.

When to combine them:

Either method can be used individually, but for compound trusses, combining them often speeds up analysis.
Sometimes, in the method of joints, you cannot find a joint with two or fewer unknown forces to start solving.

Solution in such cases:

Use the method of sections to determine some member forces.
This will create a joint with two or fewer unknowns, allowing you to continue with the method of joints.

Example

1. Determine the force in each member of the Fink truss shown in Fig.

Solution – Analysis of the Fink Truss

Truss Description

The given Fink truss is a compound truss.
Formed by connecting two simple trusses (ACL and DFL) at a common joint L and with a connecting member CD.

1. Static Determinacy Check

Number of members (m) = 27
Number of joints (j) = 15
Number of reactions (r) = 3
Static determinacy equation:

m+r=2j = 27+3=2(15)30=30 ⇒ Statically determinate

Members and reactions are properly arranged → Truss is stable and determinate.

2. Support Reactions

Applying the three equations of equilibrium to the entire truss:

Ax=0, Ay=42 kN (↑), Fy=42 kN (↑)

3. Method of Joints

Joint A

Unknown forces: F_AB and F_AI (only two unknowns).
From equilibrium (𝛴Fx and 𝛴Fy):

F_AI=93.91 kN (C)

F_AB=84 kN (T)

Joint I

Member BI is perpendicular to AI and IJ (which are collinear).
Take x-axis along AI and IJ (Direction of the colinear members) for simplification.

Joint B

Considering the equilibrium of joint B, we obtain, the following:

Using Section

Section aa to Find FCD

Reason for using Section aa:

At Joint C → 3 unknown forces (F_CD, F_CG, F_CJ)
At Joint J → 3 unknown forces (F_CJ, F_GJ, F_JK)
Too many unknowns for the method of joints to proceed directly.
Solution: Use the method of sections to cut the truss and solve for it first.

Why not start from the other end?

If starting from Joint F, similar difficulty arises at Joints D and N (also 3 unknowns each).

Procedure:

1. Cut the truss along Section aa

2. Isolate the left portion of the truss for analysis.

3. Choose a point to take moments that eliminates other unknown forces.

§ Moment center: Point L (intersection of lines of action of F_GL and F_KL

§ This choice removes F_GL and F_KL from the moment equation, leaving F_CD as the only unknown.

Joint C.

With FCD now known, there are only two unknowns, FCG and FCJ, at joint C. These forces can be determined by applying the two equations of equilibrium to the free body of joint C

Joints J, K, and G

Similarly, by successively considering the equilibrium of joints J, K, and G, in that order, we determine the following:

Joint J

Joint K

Solving these equations:

F_GK = -10.73 K = 10.73 (C)

F_KL = -77.81 K = 77.81 (C)

Joint G

Symmetry. Since the geometry of the truss and the applied loading are symmetrical about the center line of the truss (shown in Fig. above), its member forces will also be symmetrical with respect to the line of symmetry. It is, therefore, sufficient to determine member forces in only one-half of the truss.

Space Trusses

Space Trusses – Overview

Definition:

Trusses that cannot be subdivided into plane trusses due to their shape, member arrangement, or loading.
Must be analyzed as three-dimensional structures with 3D force systems.

Simplifying Assumptions (for Analysis)

Members are connected at ends by frictionless ball-and-socket joints.
All external loads and reactions act only at the joints.
The centroidal axis of each member coincides with the straight line between the centers of adjacent joints.
With these assumptions → Members carry only axial forces.

Basic Space Truss Element

Simplest internally stable space truss = Tetrahedron:

4 joints, 6 members, connected by ball-and-socket joints.
Internally stable → maintains its shape under general 3D loading at joints.

Enlarging a Space Truss

1. Start with basic tetrahedron (e.g., truss ABCD).

2. Add one new joint (E) and connect it to three existing joints (e.g., B, C, D) with three new members (BE, CE, DE).

3. Condition: New joint must not lie in the plane of the three existing joints.

4. Repeat the process to enlarge further.

Simple Space Trusses

Constructed by repeatedly enlarging the basic tetrahedron in the above way.
Each new joint → add 3 new members.
Member–Joint Relationship Formula:

m=3j−6

where:

m = total number of members
j = total number of joints (including those attached to supports)

Types of Supports for Space Trusses

Reactions in Space Trusses

1. Support Types

Common support types for space trusses are shown in Fig. 4.29.
Number and direction of reaction forces depend on how many translations the support prevents.

2. Equilibrium Requirement

For an internally stable space truss to be in equilibrium under general 3D loading:

Must have at least 6 independent reactions.
These correspond to the six equations of equilibrium:

3. Static Determinacy Conditions

Let r = number of reactions:

These conditions are necessary but not sufficient → Proper arrangement of supports is also required for geometric stability.

5. Geometric Stability Requirement

Reactions must be placed so they:

Prevent translations in all 3 directions (x,y,z).
Prevent rotations about all 3 axes.

If lines of action of all the reaction are parallel or meet at a common axis, the truss will be geometrically unstable, even if r=6.

Static Determinacy, Indeterminacy, and Instability

If a space truss contains m members and is supported by r external reactions, then the conditions of static instability, determinacy, and indeterminacy of space trusses can be summarized as follows:

Analysis of Space Trusses

1. Extending Plane Truss Methods

The method of joints and method of sections used for plane trusses also apply to space trusses.
Method of Joints (3D version):

At each joint, satisfy:

ΣFx=0, ΣFy=0, ΣFz=0

Start at a joint with no more than 3 unknown forces (must not be coplanar).
Solve the 3 unknowns using the 3 equilibrium equations.
Proceed joint by joint until all required member forces are found.

2. Force Components in Space Truss Members

For member AB:

L_AB = total length of member
xAB, yAB, zABx = projections in x, y, z directions

3. Zero-Force Members in Space Trusses

Rule 1:

If all but one of the members at a joint lie in a single plane,
and no external load/reaction is applied at that joint →
the non-coplanar member has zero force.
Example: Fig. (a) → F_AE = 0.

· The first type of arrangement is shown in Fig. (a). It consists of four members AB, AC, AD, and AE connected to a joint A. Of these, AB, AC, and AD lie in the xz plane, whereas member AE does not. Note that no external loads or reactions are applied to joint A. It should be obvious that in order to satisfy the equilibrium equation 𝛴Fy = 0, the y component of FAE must be zero, and therefore FAE = 0.

Rule 2:

If all but two members at a joint have zero force,
and no external load/reaction is applied →
unless the two remaining members are collinear,
both have zero force.

Example: Fig. (b)

At joint A, members AB, AC, AD, AE are connected.
Suppose AD and AE have already been identified as zero-force (maybe from Case 1 or symmetry).
That leaves only AB and AC.
No external load is applied at A.
Since AB and AC are not collinear (not in a straight line), they cannot balance each other in all 3 directions.

So:

∑Fy = 0 and ∑Fz = 0 force F_AC = 0.
With AC, the x-component balance ∑Fx = 0 requires F_AB = 0 as well.
“With F_AC = 0, the joint no longer has any y- or z-forces.
That leaves only AB in the x-direction.
Equilibrium requires ∑Fx = 0, but since FAB is the only x-force left, the only way for the sum to be zero is if F_AB=0

Therefore, F_AB = F_AC = 0.

4. Method of Sections

Cut through the truss to expose desired members.
Apply six equations of equilibrium:

ΣFx=0, ΣFy=0, ΣFz=0,

ΣMx=0, ΣMy=0, ΣMz=0

No more than six unknown member forces can be found from one section.

5. Practical Considerations

Manual calculations for large space trusses are very time-consuming.
In practice, computer analysis is used for large systems.
Why learn manual methods?

Builds understanding of geometry, load paths, and force distribution.
Helps validate computer results.

Example

1. Determine the reactions at the supports and the force in each member of the space truss shown in Fig.(a).

1) Static Determinacy

Data:

Members m=9
Joints j=5
Reactions r=6

Check:

m + r = 9 + 6 = 15,

3j = 3×5 =15

Since m + r = 3j and the supports/members are properly arranged, the truss is statically determinate.

2) Member Projections (Geometry Setup)

3) Zero-Force Members (by inspection at Joint D) [No external force or reaction]

At joint D: members AD, CD, DE meet.
Observation 1 (Rule 1 – out-of-plane):

AD and CD lie in the same xz-plane; DE is not in that plane.
No external load/reaction at D.
To satisfy ∑Fy = 0 (out of the xz-plane), the non-coplanar member must carry no force:

F_DE = 0 (zero-force)

Observation 2 (Rule 2 – two non-collinear members left):

With DE removed, only AD and CD remain at D.
They are not collinear, and still no external load at D.
Two non-collinear members at an unloaded joint cannot self-balance in 3D → both must be zero:

F_AD = 0, F_CD = 0

Reactions

Fig.(a)

By substituting Ay, By we obtain

By substituting By = 30 into, we obtain Ay = 12.5 k ↑

Joint A

in which the second term on the left-hand side represents the y component of F_AE. Substituting the values of y and L for member AE from Table, we write