Lecture 17 February 23, 2018

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1 Statics - TAM 20 & TAM 2 Lecture 7 ebruary 23, 208

Announcements Monday s lecture: watch for Piazza announcement over weekend for possible change Concept Inventory: Ungraded assessment of course knowledge Extra credit: Complete # or #2 for 0.

2 Announcements Monday s lecture: watch for Piazza announcement over weekend for possible change Concept Inventory: Ungraded assessment of course knowledge Extra credit: Complete # or #2 for 0.5 out of 00 pt of final grade each, or both for.5 out of 00 pt of final grade #: Sign up at CBT (2/26-3/ M-Th) #2: (4/2-4 M-Th) 50 min appointment, should take < 30 min Upcoming deadlines: Quiz 3 (2/2-23) Sign up at CBT Monday (2/26) Mastering Engineering Tutorial 7 Tuesday (2/27) PL HW 6 Thursday (3/) WA 3 See enhanced instructions

3 Chapter 6: Structural Analysis

4 Goals and Objectives Determine the forces in members of a truss using the method of joints Determine zero-force members Determine the forces in members of a truss using the method of sections

5 Recap: Zero-force members Two situations: Joint with ONLYtwo non-collinear members, no external or support reaction applied to the joint Both members are zero-force members. Why not also point E?

6 Recap: Picture of truss structure from this week s Discussion Section. Recall: a truss is composed of only slender, straight 2-force members and these members are joined by pins (pin joints only forces & no moment); so all resultant forces in the member are directed along the axis of the member and concentrated only at the end joints (Lecture 5). Neglect the weight of the truss members (or split weight and include as part of the resultant force). Note joints A and D are connected to ground by hinge or pin supports (Lecture 2). Method of Joints (Lecture 5): Use to solve for the resultant forces on one or more pin joints in the truss. Procedure: (i) Created BD of each joint assume unknown forces point in the direction of tension (or away from the pin). {When drawing forces: draw only one single resultant force per truss member, but both orthogonal components of support reaction force.} (ii) Start with joint with at least known force and -2 unknown forces (since only have 2 eqns of equilibrium for a pin joint σ x = 0 and σ y = 0). Assume no added loads on truss in this current embodiment.

8 The truss, used to support a balcony, is subjected to the loading shown. Approximate each joint as a pin and determine the force in each member. State whether the members are in tension or compression. Solution: Start by setting the entire structure into external equilibrium. Draw the BD. Equilibrium requires σ = 0 and σ M C = 0 0 : C E 0, x x x 0 : C P P 0, y 2 2 M 0 : 2aP ap ae 0. C Solving these equations gives the external reactions C 2 P P, C P P, E (2 P P ). x 2 y 2 x 2 y x

9 Next, start with a joint, draw the BD, set it into force equilibrium only, and move to the next joint. Start with joints with at least known force and -2 unknown forces. or example, start with joint A: 0 : 0, x AB 2 AD y 2 AD 0 : P 0. 2 P, 2 P P. AD AB 2 C 2 P P, C P P, E (2 P P ). x 2 y 2 x 2 Joint B: Joint C: Joint E: Joint D: only needed for check 0 : 0, x AB BC 0 : P 0. y 2 BD P, P. BC AB BD 2 0 : 2P P 0, x BC 2 CD y 2 CD 2 0 : P P (2 P P P ) 2( P P ), CD 2 2 2( P P ) (check). CD 2 0 : 0, x 2 AD 2 CD DE y 2 AD BD 2 CD 0 : 0. 2P 2( P P ) (2 P P ), DE P P 2( P P ) 0 (check)

10 Note: The checks would not have been satisfied if the external reactions had been calculated incorrectly. Note: The order in which the joints are set in equilibrium is usually arbitrary. Sometimes not all member loads are requested. If provided numerical values: P 800 lb P2 0 AB BC AD BD 2P 30 lb (C) 2 0 2( P P ) 30 lb (T) CD 2 (2 P P ) 600 lb (C) DE P 800 lb (T) P 800 lb (T) P 2 Note that, in the absence of P 2, member BD is a zero-force member

E Equilibrium of the entire structure requires: y x Joint E: Joint D: 0 : E 0, x 5 29 E DE x y 2 29 E Ey E 29 29 E 2 y 2 0 : 0. 5 5 DE 29 E x 29 A.

11 Determine the force in member G of the truss and state if the member is in tension or compression. Solution: Draw a free-body diagram of the entire truss 0 : 2.5 E 0, x 0 : 3 3 A E 0, y y y M 0 : 40A 4(.5) 2(2) 20(3) 0(3) 0. E Equilibrium of the entire structure requires: y x Joint E: Joint D: 0 : E 0, x 5 29 E DE x y 2 29 E Ey E E 2 y 2 0 : DE 29 E x 29 A.5 kips, E 3.5 kips, E 4.5 kips. y x y (4.5) 2.2 kips (C), E ( 2.2) ( 3.5) 7.75 kips (T). 0 : 0, x CD DE 0 : 0. y D CD D 0. DE 7.75 kips (T), Note: Member D turned out to be a zero-force member. Under what conditions would this result not be true?

Joint : 0 : 0, 5 5 5 x 29 G 29 C 29 E 2 2 2 y 29 G 29 C D 29 E 0 : 3 0. G C E 29 3. G C E 2 D 29 G E 2 2 3 D, 29 4 2.

12 Joint : 0 : 0, x 29 G 29 C 29 E y 29 G 29 C D 29 E 0 : 3 0. G C E G C E 2 D 29 G E D, kips (C). Note: Nine scalar equations of equilibrium were needed to obtain this answer. Might there be a shorter way?

13 Method of sections Determine external support reactions Cut the structure at a section of interest into two separate pieces and set either part into force and moment equilibrium (your cut should be such that you have no more than three unknowns) Determine equilibrium equations (e.g., moment around point of intersection of two lines) Assume all internal loads are tensile.

14 Method of sections Determine equilibrium equations (e.g., moment around point of intersection of two lines) Assume all internal loads are tensile.

15 Determine the force in member BC of the truss and state if the member is in tension or compression.

16 Determine the force in member BC of the truss and state if the member is in tension or compression

17 Determine the force in members OE, LE, LK of the Baltimore truss and state if the member is in tension or compression. Solution: Use method of sections, since cutting LK, LE, OE, and DE will separate the truss into two pieces. Note that LE is a zero-force member. Draw freebody diagram of entire structure, and set into external equilibrium: 0 : A 0, x 0 : A I , y y y M 0 : 2 a(2) 3 a(2) 4 a(5) 5 a(3) 8aI 0. A A 0, A kn, I kn. x y y x y Normally, introducing four unknowns would make the problem intractable. However, LE is a zero-force member. Set either remaining section into equilibrium. Here, there is no real preference, but the right half will be fine

18 0 : 0, x DE 2 EO KL y 2 EO y 0 : I 0, M 0 : 0(5) a( 3) 4aI 2a 0. E y KL DE EO 7.38 kn (T), 0 (zero-force), EL 3.36 kn (T), 9.75 kn (C). KL

Lecture 17 February 23, 2018

Lecture 17 February 23, 2018 Statics - TAM 20 & TAM 2 Lecture 7 ebruary 23, 208 Announcements Monday s lecture: watch for Piazza announcement over weekend for possible change Concept Inventory: Ungraded assessment of course knowledge