This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CIVL 31 1 Truss Analysis by Method of Joints What makes a truss? members straight
" pinned" joints = two force member loads supported at joints In a twoforce member the alignment of the internal force is known; it is the same as the alignment of
the member. Thus, each member contains only one unknown; the magnitude (and associated sign,
tension or compression) of the internal force. Shown below is a typical plane truss and a freebody diagram for a typical member, AB. Since standard
convention is tension +, the internal force is shown as a tension force. Notice that since the force and
member are each at the same alignment, similar triangles may be used to relate the two components of
force to the components of the member’s length with a simple ratio! In the Method of Joints, freebody diagrams of the joints are repetitively used to calculate the unknown
internal member forces. For a plane truss, as long as there are no more than two unknown forces at a joint it is possible to solve for these member forces using two equations of statics. It is extremely convenient to write these equilibrium equations not in terms of the internal forces
directly, but in terms of the components of these forces! Do not use angles, sines or cosines to relate
these components to each other and to their resultant. Instead, use the components of length. For example, consider a FBD of joint 13 of the truss shown previously. That FBD and the associated
equilibrium conditions would be developed as shown below. AB \ ABV EFV=O: ABV=5kips ‘~ L 20'
ABH “ AB =AB [—H! =(5) — =10kips
O H V [L J 10!
BC V
Skips 2 PH = 0: BC = —AB,, = —10kips With practice, it becomes no longer necessary to draw these FEDS or even to write out the equilibrium
equations. Instead, use the virtual—reality imaging capabilities of your own biological neural processor! CIVL 311 Truss Analysis by Method of Sections What & How: With this technique, a section line is used to completely separate the truss into sections.
The contribution of members cut by the section line must be represented by force vectors. Again, the
alignment of these vectors are known; only their numerical values need be calculated. (Hint: Make sure
your section line cuts only members; do not attempt to cut through a joint with a section!) In the case of a plane truss, as long as the unknown internal member forces that are exposed by the
section line are n_ot convergent at a point (i.e., concurrent) three equations of static equilibrium may be
written. Consequently, it is possible to solve for up to three internal member forces with a single section. Helpful Hints: The selected equilibrium equations may be either written in terms of force or in terms of
moment. Moment equations can be very powerful since judicious selection of the point about which
moments are calculated will help eliminate unknowns from the equation. Also, just like in the Method of
Joints, it is essential that these equations, whether force summation or moment summation, be written in
terms of the components of the unknown member forces! Use the member geometry to relate the
individual components. (Note: Components need not be vertical and horizontal.) A complete and efﬁcient truss analysis may utilize both the joint and section methods. As a rule of
thumb, the Method of Joints is used when all or most member forces are desired and Method of Sections
is used when only a few member forces are needed. (However, all rules of thumb are invalid at least
fifty percent of the time.) Example: Here is a plane truss. Let’s use Method of Sections to calculate the internal force supported by members AB and AD.
',.'. A H 20' H4 20' H Section 11, Right FBD: 2 [email protected] = 0: 20kips(20') — A3,,(5') = 0 20 kips
ABH‘.._ ABH = 80 kips (T)
AD i ABJiUB =20kis T
v L20.J( H) p ( ) DE AB: J802 +202 =82.5kips (T)
2F, = o — 20kips+ABV+ADV =0
A1), = 0 Principle of Transmissibility: I chose to break force AB into components at joint B. I could also have
moved force AB along its line of action to joint C, and broken it into components there. In which case, summing moments at D would have provided the vertical component of AB! ...
View
Full
Document
This note was uploaded on 01/07/2012 for the course CIVL 311 taught by Professor Mills during the Spring '09 term at CSU Chico.
 Spring '09
 Mills

Click to edit the document details