Trusses WORKSHEET10 to answer just click on the button or image related to the answer

Question 1 what is a truss? a triangular arrangement of members with rigid joints a a triangular arrangement of members with pinned joints b a beam with holes c same as a frame d

Question 2a which members resist the bending moments? top chords a bottom chords b vertical web members c diagonal web members d a & b e c & d f a & c g b & d h in a parallel chord truss

Question 2b which members resist the shear forces? top chords a bottom chords b vertical web members c diagonal web members d a & b e c & d f a & c g b & d h in a parallel chord truss

Question 2c which members resist the compressive forces from the bending moment? top chords a bottom chords b middle top chords c middle bottom chords d in a parallel chord truss

Question 2d where does the maximum compressive force from bending occur? at a (middle bottom chords) a at b (middle top chords) b at c (end diagonal web members) c at d (vertical web members) d in the parallel chord truss shown a bc d

Question 2e where does the maximum tensile force from bending occur? in the parallel chord truss shown a bc d at a (middle bottom chords) a at b (middle top chords) b at c (end diagonal web members) c at d (vertical web members) d

Question 2f where does the maximum shear force occur? in the parallel chord truss shown a bc d at a (middle bottom chords) a at b (middle top chords) b at c (end diagonal web members) c at d (vertical web members) d

Question 3a where does the maximum compressive force from bending occur? at a (end top chords) a at b (end bottom chords) b at c (middle diagonal web members) c at d (middle vertical web members) d in the triangular truss shown a bc d

Question 3b where does the maximum tensile force from bending occur? at a (end top chords) a at b (end bottom chords) b at c (middle diagonal web members) c at d (middle vertical web members) d in the triangular truss shown a bc d

Question 3c where does the maximum shear force occur? at a (end top chords) a at b (end bottom chords) b at c (middle diagonal web members) c at d (middle vertical web members) d in the triangular truss shown a bc d

Question 4a what is the force in the horizontal component, H? 10.0 kN a 14.1 kN b 7.1 kN c given the force shown 45 o 10kN H V

Question 4b what is the force in the vertical component, V? 10.0 kN a 14.1 kN b 7.1 kN c given the force shown 45 o 10kN H V

Question 5a what is the force in the horizontal component, H? 12.0 kN a 5.2 kN b 7.1 kN c given the force shown H V 30 o 6kN

Question 5b what is the force in the vertical component, V? 8.0 kN a 3.0 kN b 12.0 kN c given the force shown H V 30 o 6kN

Question 6 what is the resultant force? 10 kN at to the horizontal a 7 kN at to the horizontal b 10 kN at to the horizontal c given the forces shown 6kN 8kN use the parallelogram of forces or the triangle of forces

Question 7 when would we use the method of sections? to find the forces in all the members a to find the force in only certain members b because it’s the easiest c when analysing a truss

Question 8a what do we do first? make a cut through CE a take moments b find the reactions c using the Methods of Sections to find the value of the force in the member CE 1kN 2kN 1kN 45 o A BC D E F 4 3m 3m

Question 8b what are the reactions? R L = 8 kN R R = 8 kN a R L = 4 kN R R = 4 kN b R L = 8 kN R R = 4 kN c using the Methods of Sections to find the value of the force in the member CE 1kN 2kN 1kN 45 o A BC D E F 4 3m 3m RLRL R

Question 8c what do we do next? make a cut through CE a take moments b make a freebody c using the Methods of Sections to find the value of the force in the member CE once we have the reactions 1kN 2kN 1kN 45 o A BC D E F 4 3m 3m R L = 4kN R R = 4kN

Question 8d what do we do next? add in all the forces a use the equations of static equilibrium b take moments c using the Methods of Sections to find the value of the force in the member CE once we have made the cut 1kN 2kN 1kN 2kN A BC D E F 4 kN 3m

Question 8e what do we do next? eliminate some of the unknowns a use the equations of static equilibrium b take moments c using the Methods of Sections to find the value of the force in the member CE once we have added in forces CE, CF and DF 1kN 2kN 1kN 2kN A BC D E F 4 kN 3m

Question 8f which equation is the most useful here? ΣV = 0 a ΣH = 0 b ΣM = 0 c using the Methods of Sections to find the value of the force in the member CE given ΣV = 0, ΣH = 0, ΣM = 0 1kN 2kN 1kN 2kN A BC D E F 4 kN 3m

Question 8g what does that mean? we will take moments about a point a we will calculate clockwise moments b we will calculate anticlockwise moments c using the Methods of Sections to find the value of the force in the member CE given that we will use ΣM = 0 1kN 2kN 1kN 2kN A BC D E F 4 kN 3m

Question 8h about which point? A a C b using the Methods of Sections to find the value of the force in the member CE given that we will take moments about a point F c D d 1kN 2kN 1kN 2kN A BC D E F 4 kN 3m

Question 8i why do we take moments about F? seems right a it is perpendicular to CE b using the Methods of Sections to find the value of the force in the member CE eliminates the forces CF and DF c 1kN 2kN 1kN 2kN A BC D E F 4 kN 3m

Question 8j what is the value of the force in CE? 4 kN compression a 4 kN tension b using the Methods of Sections to find the value of the force in the member CE 12 kN tension c 12 kN compression d 1kN 2kN 1kN 2kN A BC D E F 4 kN 3m

1kN 2kN 1kN 2kN A BC D E F 4 kN 3m Question 8k where would we take moments? using the Methods of Sections to find the value of the force in the member DF A a C b E c D d

next question enough ! a truss is a linear arrangement of short members, arranged to form triangles. The joints are pinned and the loads should be placed at the joints so that the members are only in tension or compression.

let me try again let me out of here rigid joints make a frame not a truss

let me try again let me out of here in some way you can think of trusses as beams with holes, i.e. the chords are the flanges and the web members are similar but trusses have specific arrangements to avoid bending moments.

trusses do not behave like frames frames have rigid joints trusses avoid bending moments let me try again let me out of here

next question enough ! the chords (all of them) resist the bending moments

partly right let me try again let me out of here

the web members DO NOT resist the Bending Moments let me try again let me out of here

next question enough ! the web members (all of them) resist the shear forces

the chords resist the Bending Moments NOT the Shear Forces let me try again let me out of here

partly right let me try again let me out of here

next question enough ! the top chords (all of them) resist the compressive forces from the bending moment – just as the top part of a beam

think of a beam under bending let me try again let me out of here

partly right let me try again let me out of here

next question enough ! the top chords resist the compressive forces from the bending moment but the maximum forces occur in the middle (as in a simply supported beam)

didn’t we say that it is the top chords that resist the compressive forces let me try again let me out of here

didn’t we say that it is the top chords that resist the compressive forces the web members do not resist the bending forces let me try again let me out of here

next question enough ! the bottom chords resist the tensile forces from the bending moment but the maximum forces occur in the middle (as in a simply supported beam)

didn’t we say that the top chords resist the compressive forces let me try again let me out of here

didn’t we say that the web members do not resist the bending forces let me try again let me out of here

next question enough ! the web members resist the shear forces but the maximum forces occur at the end (as in a simply supported beam with a UDL)

didn’t we say that the chords resist the forces from the bending moment and that it is the web members which resist the shear forces let me try again let me out of here

the maximum shear forces do occur at the ends just like in a simply supported beam with a UDL but not usually in the vertical members let me try again let me out of here

next question enough ! unlike in a parallel chord truss, the maximum compressive force in a triangular truss occurs near the supports

didn’t we say that the top chords resist the compressive forces let me try again let me out of here

didn’t we say that the web members do not resist the bending forces let me try again let me out of here

next question enough ! unlike in a parallel chord truss, the maximum tensile force in a triangular truss occurs near the supports

didn’t we say that the top chords resist the compressive forces let me try again let me out of here

didn’t we say that the web members do not resist the bending forces let me try again let me out of here

next question enough ! unlike in a parallel chord truss, the maximum shear force in a triangular truss occurs in the middle

didn’t we say that the chords resist the forces from the bending moment let me try again let me out of here

the maximum shear forces do occur in the middle but not usually in the vertical member let me try again let me out of here

next question enough ! H = 10 cos 45 = 7.07 kN to the right 45 o 10kN H V

let me try again let me out of here How did you get that? what is the expression for the horizontal component? Look at your notes!

next question enough ! V = 10 sin 45 = 7.07 kN downwards 45 o 10kN H V

let me try again let me out of here How did you get that? what is the expression for the vertical component? Look at your notes!

next question enough ! H = 10 cos 30 = 5.2 kN to the right H V 30 o 6kN

let me try again let me out of here How did you get that? what is the expression for the horizontal component? Look at your notes!

next question enough ! V = 10 sin 30 = 3.0 kN downwards H V 30 o 6kN

let me try again let me out of here How did you get that? what is the expression for the vertical component? Look at your notes!

next question enough ! R = √( ) = 10.0 kN 6kN 8kN RkN Ø Ø = arctan(6/8) = tan = o 6 = 3 x 2 8 = 4 x 2 10 = 5 x 2 did you notice that this is a right-angled triangle?

let me try again let me out of here How did you get that? have another look at the triangle formed. How do you get the resultant? How do you calculate the angle?

next question enough ! yes, because we just cut a section through the member we are interested in. If we needed all the member forces we would use the Method of Joints

let me try again let me out of here we could, but that’s not the best use for this method

let me try again let me out of here you were kidding, …. weren’t you?

next question enough ! yes, we must first find all the forces acting and that includes the reactions

let me try again let me out of here we need to do something first

let me try again let me out of here about where would we take moments?

next question enough ! the total downward loads = 8 kN so the total upward loads = 8kN (ΣV = 0) the loads are symmetrical – so the reactions must be equal

let me try again let me out of here what’s the sum of the downward forces? ΣV = 0, so how can the sum of the upward forces be bigger?

next question enough ! we cut through the member whose force we wish to find

where would we take moments? let me try again let me out of here

how do we do that? How do we isolate what we want? let me try again let me out of here

next question enough ! yes, before we can do anything we must add in all the forces the reaction and forces CE, CF and DF 1kN 2kN 1kN 2kN A BC D E F 4 kN 3m

we need to know something first? let me try again let me out of here

next question enough ! you’ve got it !!

let me try again let me out of here first things first

let me try again let me out of here first things first we may, but why do we do that?

next question enough ! that’s it exactly

let me try again let me out of here how would that help?

next question enough ! you’ve got it !!

let me try again let me out of here first things first we may, but why do we do that?

next question enough ! that’s it exactly

let me try again let me out of here the idea is to make the calculations simpler

let me try again let me out of here what would that do to the forces CE and CF ? Isn’t CE the force that we are trying to find?

let me try again let me out of here what do we do about the force CF?

next question enough ! you’ve got it !! Since we are after the force CE we want to eliminate the other two forces, CF and DF. To do that we take moments about the point through which they both go through, i.e. point F since the moment of a force passing through a point is 0

let me try again let me out of here you can do better than that remember that the moment of a force passing through a point is 0

So is point D. But what has that to do with it? let me try again let me out of here

next question enough ! taking moments about point F, we get: clockwise moments = anticlockwise moments 4 x 6 + CE x 3 = 1 x x 3 3CE = CE = - 4 Since the answer is negative, CE is in compression (as we would expect) 1kN 2kN 1kN 2kN A BC D E F 4 kN 3m

are the top chords in tension or compression? let me try again let me out of here

It’s not correct. Try again. Just get your clockwise and your anticlockwise moments correct let me try again let me out of here

that’s it – all done Yes! We want to eliminate forces CE and CF. So we take moments about C I think you’ve got it !

let me try again let me out of here the idea is to make the calculations simpler

let me try again let me out of here what do we do about the force CF?

let me try again let me out of here what would that do to the force DF ? Isn’t DF the force that we are trying to find?