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Trusses are used commonly in Steel buildings and bridges.

**Definition: A truss is a structure that consists of**

- All straight members
- Connected together with pin joints
- Connected only at the ends of the members
- All external forces (loads & reactions) must be applied only at the joints.
- Trusses are assumed to be of negligible weight (compared to the loads they carry)

**Types of Trusses**

Typical Bridge Trusses

- D
_{S}= m+r_{e}– 2j where, D_{S}= Degree of static indeterminacy m = Number of members, r_{e}= Total external reactions, j = Total number of joints - D
_{S}= 0 ⇒ Truss is determinate - If D
_{se}= + 1 & D_{si }= –1 then D_{S}= 0 at specified point. - D
_{S}> 0 ⇒ Truss is indeterminate or dedundant.

- Conditions of equilibrium are satisfied for the forces at each joint
- Equilibrium of concurrent forces at each joint
- Only two independent equilibrium equations are involved

**Steps of Analysis**

- Draw Free Body Diagram of Truss
- Determine external reactions by applying equilibrium equations to the whole truss
- Perform the force analysis of the remainder of the truss by Method of Joints

**Example 1: **Determine the force in each member of the loaded truss by Method of Joints

**Solution:**

[ΣF_{y} = 0] 0.866BC - 0.866(34.6) - 20 = 0

Cd = 57.7 kn T

[ΣF_{x} = 0 CE - 17.32 - 0.5(34.6) - 0.5(57.7) = 0

CE = 63.5 kN C

[ΣF_{y} = 0] 0.866DE = 10 DE = 11.55 kN C

and the equation ΣF_{x }= 0 checks.

**Truss Member Carrying Zero forces**

**(i)** M_{1}, M_{2}, M_{3} meet at a joint M_{1} & M_{2} are collinear ⇒ M_{3 }carries zero force where M_{1}, M_{2}, M_{3} represents member.

**(ii) **M_{1} & M_{2} are non collinear and F_{ext} = 0 ⇒ M_{1} & M_{2} carries zero force.

- If only two non-collinear members form a truss joint and no external load or support reaction is applied to the joint, the two members must be zero force members
- If three members form a truss joint for which two of the members are collinear, the third member is a zero-force member provided no external force or support reaction is applied to the joint.

**Method of Section**

- It can be used to determine three unknown member forces per FBD since all three equilibrium equations can be used
- Equilibrium under non-concurrent force system
- Not more than 3 members whose forces are unknown should be cut in a single section since we have only 3 independent equilibrium equations

**Principle**

- If a body is in equilibrium, then any part of the body is also in equilibrium.
- Forces in few particular member can be directly found out quickly without solving each joint of the truss sequentially
- Method of Sections and Method of Joints can be conveniently combined
- A section need not be straight.
- More than one section can be used to solve a given problem

**Example 2: The truss in Fig given below is pinned to the wall at point F, and supported by a roller at point C. Calculate the force (tension or compression) in members BC, BE, and DE.**

**Solution: **From section to the left of a-a

[ΣFV = 0

5/√29 F_{BE} = 80 + 60

F_{BE} = 150.78 kN tension

ΣM_{E }= 0

5F_{BC }= 6(80)+2(60)

F_{BC }= 120 kN compression

ΣM_{B }= 0

5F_{DE }= 4(80)

F_{DE }= 64 kN Tension

**(i) **Final force in the truss member

sign convn → +ve for tension, –ve for compression

where,

S = Final force in the truss member

K = Force in the member when unit load is applied in the redundant member

L = Length of the member

A = Area of the member

E = Modulus of elasticity

P = Force in the member when truss become determinate after removing one of the member.

P = Zero for redundant member.**Lack of Fit in Truss**

Q = Force induce in the member due to that member which is 'Δ' too short or 'Δ' too long is pulled by force 'X'.

**Deflection of Truss**

Where, y_{C} = Deflection of truss due to effect of loading & temp. both.

If effect of temperature is neglected then

α = Coefficient of thermal expansion

T = Change in temperature

T = +ve it temperature is increased

T = -ve it temperature is decreased

P & K have same meaning as mentioned above.

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