SOLIDS AND STRUCTURES- FRAMES, Study notes of Engineering

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Peer Assisted Learning (PAL)

• Get support from other students in years above you

i.e. student led

• Fridays 12-1pm in ALL 303
• Not compulsory
• Covers Solids, Fluids, Materials and Thermo
• Come along to share with and learn from others
• Promotes deeper learning

APPLICATIONS

Trusses are commonly used to support roofs. For a given truss geometry and load, how can you determine the forces in the truss members and thus be able to select their sizes? A more challenging question is, that for a given load, how can we design the trusses’ geometry to minimize cost?

APPLICATIONS (continued)

Trusses are also used in a variety of structures like cranes and the frames of aircraft or this space station. How can you design a lightweight structure satisfying load, safety, cost specifications,

A simple truss is a planar truss which begins with a triangular element and can be expanded by adding two members and a joint. For these trusses, the number of members (M) and the number of joints (J) are related by the equation M = 2 J – 3.

ANALYSIS & DESIGN ASSUMPTIONS

When designing the members and joints of a truss, first it is necessary to determine the forces in each truss member. This is called the force analysis of a truss. When doing this, two assumptions are made:

1. All loads are applied at the joints. The weight of the truss members is often neglected as the weight is usually small as compared to the forces supported by the members.

2. The members are joined together by smooth pins. This assumption is satisfied in most practical cases even where the joints are formed by bolting the ends together. With these two assumptions, the members act as two-force members. They are loaded in either tension or compression. Often compressive members are made thicker to prevent buckling.

METHOD OF JOINTS (or Equilibrium at a point)

A free-body diagram of Joint B

500 N

B

F

BC

(comp)

F

AB

(tension)

3. Apply the scalar equations of equilibrium, FX = 0 and FY = 0, to determine the unknown(s). If the answer is positive, then the assumed direction (tension) is correct, otherwise it is in the opposite direction (compression).
4. Repeat steps 2 and 3 at each joint in succession until all the required forces are determined.

ZERO-FORCE MEMBERS

If a joint has only two non-collinear members and there is no external load or support reaction at that joint (e.g. A and D), then those two members are zero- force members. In this example members DE, DC, AF, and AB are zero force members.

You can easily prove these results by applying the equations of equilibrium to joints A and D. Zero-force members can be removed (as shown in the figure) when analyzing the truss.

ZERO – FORCE MEMBERS (continued)

If three members form a truss joint for which two of the members are collinear (e.g. joints C and D) and there is no external load or reaction at that joint, then the third noncollinear member is a zero-force member, e.g. AC and AD. Again, this can easily be proven. The zeroforce members can also be removed, as shown, on the left, for analyzing the truss further.

WORKED EXAMPLE

Given: Loads as shown on the truss Find: The forces in each member of the truss. Plan:

1. Check if there are any zero-force members.
2. First analyze pin D and then pin A.
3. Note that member BD is a zero-force member. FBD = 0
4. You don’t have to find the external reactions before solving this problem? This is not the case for every problem.

D

45

WORKED EXAMPLE (continued)

450 kN 45

FAD FCD

FBD of pin D y (+)

• FX = – 450 + FCD cos 45° – FAD cos 45° = 0 (1)
• FY = – FCD sin 45° – FAD sin 45° = 0 (2) x (+) FCD = 318 kN (Tension) or (T) and FAD = – 318 kN (Compression) or (C) Substitute (2) FAD = - FCD into (1)

CONCEPT QUIZ

1. Truss ABC is changed by decreasing its height P from H to 0.9 H. Width W and load P are kept the same. Which one of the following statements is true for the revised truss as compared to the original truss? A) Forces in all its members have decreased. C) Forces in all its members have remained the same. D) None of the above. B) Forces in all its members have increased.

H

A

B

C

W

Equilibrium at A P F 2sinθ P A P A

Clifford – Intro to ME:

Part 1 Unit 1 Solid

Mechanics

Hibbeler – Statics &

Mechanics of

Materials

Blockley – Structural

Engineering: A Very

Short Introduction

p.7 pin-jointed str. to p.9, & p. Ch.5 pp.241- 252 -

Self-study:

• Try self-study problems (at end of this PPT)

• Watch video – pin-jointed frame (on Canvas)

BREAK Given: Loads as shown on the truss and ignoring selfweight Find: Determine the force in all the truss members (do not forget to mention whether they are T or C).