WORKSHEET 6 TRUSSES

Q1 When would we use a truss? (a) long spans, loads not too heavy (b) when want to save weight (d) when want light appearance (c) when have plenty of depth 2/24

Q2 When would we not use a truss? (a) don’t have the depth (b) very large loads (d) can’t provide lateral support if needed (c) don’t want fussy appearance 3/24

Q3 What are the main characteristics of trusses? (a) assembly of (short) linear members (b) members connected to form triangles (c) joints pinned (d) loads applied at panel points (joints) (d) members carry only tension or compression forces 4/24

Q4 What materials are trusses most commonly made of? (a) steel (b) timber (c) concrete (very occasionally) 5/24

Q5 How are the joints commonly made? (a) in timber trusses (b) in steel trusses (i) gusset plates (i) using gangnail joints - light timber (ii) using gusset plates (iii) overlapping/double members and bolts (iv) concealed plates - bolts (ii) welded joints 6/24

Q6 What do the following do? (a) the chords (b) the web members (i) resist the shear forces (ii) top chords take the compressive forces (in a triangular truss, the top chord also resists shear) (iii) bottom chords take the tensile forces (i) chords resist the bending moment 7/24

Q7 What are the common span-to-depth ratios for: (a) timber beams? (b) trusses? 5-10: :1 8/24

Q8 In the truss below show where: (a) the maximum compressive force in a chord occurs (b) the maximum tensile force in a chord occurs (c) the maximum shear force in a web member occurs (c) (a) (b) 9/24

Q9 In the truss below show where: (a) the maximum compressive force in a chord occurs (b) the maximum tensile force in a chord occurs (c) the maximum shear force in a web member occurs (c) (a) (b) 10/24

Q10 (a) (i) the horizontal component? H H = 10 cos 45 = 7.07 kN to the right V (ii) the vertical component? V = 10 sin 45 = 7.07 kN down (a) given the force shown, what is 45 o 10kN 11/24

Q10 (b) (i) the horizontal component? H H = 6 cos 30 = 5.2 kN to the right V (ii) the vertical component? V = 6 sin 30 = 3.0 kN up (b) given the force shown, what is 30 o 6kN 12/24

Q10 (c) R =  ( ) 8kN 6kN (c) given the two forces shown, use the parallelogram of forces or the triangle of forces to find the resultant force 8kN 6kN 8kN R ØoØo Tan Ø = 6 / 8 = 0.75 Ø = o R ØoØo = 10 kN at o 13/24

Q11 Analyze the truss shown below using the Methods of Joints First find the reactions. Using symmetry R1 = R2 = 8/2 = 4kN Assume all forces are in tension - i.e. away from the joint 1kN 2kN 1kN 45 o A BC D E F R1=4kN R2=4kN 4 3m 3m 14/24

Q11 (cont1.) start at joint A (support) A AB AD 4 AB = -4.0 (compression) AD = 0 B BA 4 BC BD 1 vertically: 4 + AB = 0 horizontally: next at joint B. BD = (tension) BC = -3.0 (compression) vertically: BDsin45 o = 0 horizontally: BC cos45 o = 0 next at joint D. DC = -3.0 (compression) DF = +3.0 (tension) vertically: 4.24sin45 o + DC = 0 horizontally: 0 +DF cos45 o = 0 D DC DF DA (0) DB (4.24) 15/24

Q11(cont2.) next at joint C CF = (tension) vertically: CFsin45 o = 0 horizontally: 3 +CE +1.41cos45 o = 0 CE = -4.0 (compression) next at joint E. EF = -2.0 (compression) vertically: 2 + EF = 0 C CB 3 CE CF CD 3 2kN E EC 4 EG 4 EF 2kN That completes half the truss. The other half is the same by symmetry 16/24

Q11 (cont3.) 1kN 2kN 1kN A BC D E F Note that the biggest chord forces are near the middle and the biggest web forces are near the ends Always true for parallel chord trusses with fairly uniform loading 17/24

Q12 For the same truss use the Methods of Sections to find the forces in members BD, CE and DF First find the reactions. Using symmetry R1 = R2 = 8/2 = 4kN Assume all forces are in tension - i.e. away from the joint 1kN 2kN 1kN 45 o A BC D E F R1=4kN R2=4kN 4 3m 3m 18/24

Q12(cont1.) Make a cut through the truss passing through the member wanted, BD 1kN A BC D 4 Consider the left part of the cut as a freebody Mark all the forces on it - including those members cut off the freebody is in equilibrium Using  V = BD sin45 o = 0 BD = 4.24 (tension) 19/24

Q12(cont2.) Make another cut through the truss passing through the other members wanted, CE and DF Mark all the forces on it - including those members cut off the freebody is in equilibrium Using  M = 0 about F - consider all the forces on the left 4 x 6 - 1x x 3 +CE x 3 = 0 CE = -4.0 (compression) 1kN 2kN A BC D E F 4 3m Using  M = 0 about C 4 x 3 - 1x 3 - DF x 3 = 0 DF = +3.0 (tension) Note that we have found only the members we wanted 20/24

Q13 For the same truss use the Graphical Method to construct a Maxwell diagram and find the forces in the members First find the reactions. Using symmetry R1 = R2 = 8/2 = 4kN 1kN 2kN 1kN 45 o R1=4kN R2=4kN 4 3m 3m 21/24

Q13(cont1.) First annotate using Bow’s Notation (label spaces between members and forces) First find the reactions. Using symmetry R1 = R2 = 8/2 = 4kN 1kN 2kN 1kN a bc e d f R1=4kN R2=4kN 4 3m 3m g h i j kl m n 0 22/24

Q13(cont2.) Next select a scale and draw a line representing all the loads and reactions scale for forces c d f a b g all the loads are vertical - so is the line the line is the line a,b,c,d,e,f (ag, gf) ab = 1, bc = 2,...etc ag = 4, gf = 4 e begin with a zone near a reaction, e.g h ah is vertical and gh is horizontal - meet at g (h & g at same point) hg is 0 h take next zone - i bi is horizontal and ih is at 45 o - draw these lines they meet at i i Now ij is vertical and jg is horizontal. This locates j j Proceed to k in same way and half the truss is solved k n l m o Measure all lines - this gives the force in each member need to use a special convention to determine tension or compression 23/24

Q14 When would you use: (a) the Method of joints? (b) the Method of Sections? When want to know only a few member forces e.g. end diagonals and middle chords When want to do detailed design and need to know all the forces in all members (b) the Graphical Method When don’t have a calculator or don’t want to calculate 24/24