truss examples

Examples

EX (1):

Determine the force in each member of the truss, and state if the members are in tension or compression.

Solution

Free Body Diagram:

First, we should calculate reactions at A and B:

∑MA =0

(By ( )2( + )900( )2( + )600( )4 = )0

.. .By = -2100 N = 2100N

∑Fy =0

Ay - By = 0 Ay = By

Ay = -2100N = 2100 N

∑Fx = 0

Ax – Bx + 900+ 600 = 0 Ax – Bx = -1500 ----------(1)

Node A:

Ax = 0

F1 = 2100 N (Tension)

By substituting in eqn 1, we get:

Bx =1500 N

Node D:

1

√5 F6 + 600 = 0 F6 = -1341.6 N

F6 = 1341.6 N ( compression )

F2 - 2

√5 (1341.6) = 0

(Tension)F2 = 1200 N

Node c:

CE is a zero force member,so F3 = 0 and F6 = F5 = 1341.6 N (compression)

Node E:

F4 sin45 = -900

F4 = -1272.8 N = 1272.8 (compression)

EX(2):

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. Set P1 = 3 KN, P2 = 2 KN.

Solution

Free Body Diagram:

∑MC =0

2(1 + )3(2 - )Ex (1) = 0 Ex = 8 KN

∑Fx = 0

Cx - Ex = 0 Cx = Ex

Cx = 8 KN

∑Fy = 0

Cy + Ey - 2 – 3 = 0 Cy + Ey = 5---------(1)

Node E:

Ey = 0

F1 = -8 F1 = 8 KN (compression)

By substituting in eqn 1, we get:

Cy = 5 KN

Node C:

F2 sin45 = 5 F2 = 5√2 = 7.07 KN ( Tension )

F2 cos45 + F3 = 8 5√2cos45 + F3 = 8

F3 = 3 KN (Tension)

Node B:

F5 = 3 KN (Tension)

F4 = 2 KN (Tension)

Node A:

∑Fy = 0

3 + F6 sin45 = 0

F6 = 3√2 = 4.24 KN (Tension)

EX(3):

Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 2 KN and

P2 = 1.5 KN.

Solution

Free Body Diagram:

∑MA =0

(1.5( )6(+)2( )3 + )Ex (2√3) = 0

Ex = - 4.33 KN = 4.33 KN

∑Fx = 0

Ax + (- 4.33) = 0 Ax = 4.33 KN

∑Fy = 0

Ay + Ey = 3.5 ……………..(1)

Node C :

F2 sin30 = 1.5 F2 = 3 KN (Tension)

F1 + 3cos30 = 0 F1 = - 1.5√3 = -2.6 KN=2.6 KN (compression)

Node D:

F3 = 2 KN (Tension)

F4 = - 2.6 KN =2.6 KN (compression)

Node E:

F5 cos30 + 4.33 - 2.6 = 0

F5 = - 2 KN = 2 KN (compression)

Ey + (-2) sin30 = 0 Ey = 1 KN

By substituting in eqn(1), we get

Ay + 1 = 3.5 Ay = 2.5 KN

Node A:

F6 cos60 = 2.5

F6 = 5 KN (Tension)

EX(4):

Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 4 KN and P2 = 0.

Solution

Free Body Diagram:

∑MA =0

Ey (4) – (4) (2) = 0 Ey = 2 KN

∑Fy = 0

Ay + 2 = 0 Ay = -2 KN = 2 KN

∑Fx = 0

Ex - Ax = 0 Ex = Ax ---------(1)

FD & BG are zero force members, so

F9 = 0 and F2 = 0 and as a result

AG & GC & FC & FE became zero force members,so

F1 = F3 = F10 = F8 = 0

Node A:

(0.75/1.25)F4 = 2

F4 = 10/3 = 3.33 KN (Tension)

Ax = (1/1.25) F4 = 8/3 = 2.67 KN

By substituting in eqn (1), we get

Ex = 2.67 KN

Node B:

F4 = F5 = 3.33 KN (Tension)

Node F:

(0.75/1.25 )F7 = 2

F7 = 10/3 =3.33 KN (Tension)

Node D:

F7 = F6 = 3.33 KN (Tension)

EX(5):

Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 1200 N and

P2 = 1500 N.

Solution

Free Body Diagram:

∑Mc =0

1.5 Bx – 500(3.6) =0

Bx = 1200 N

∑Fx =0

Cx +1200 +1200 =0

Cx = - 2400 N

∑Fy =0

Cy – 500 = 0 Cy = 500 N

Node D:

(1.5/3.9)F2 = - 500 F2 = - 1300 N = 1300 N (compression)

F1 + (3.6/3.9)F2 = 1200 F1 = 2400 N (Tension)

AB & AC are zero force members, so

F3 = F5 = 0

Node B:

F4 = -500 N = 500 N (compression)

EX(6):

Determine the force in each member of the truss and state if the members are in tension or compression.

Solution

Free Body Diagram:

∑Mc =0

3(1.2 + )4.5(3.6 - )Ay(2.4) = 0

Ay = 8.25 KN

∑Fx =0 Cx = 0

∑Fy =0

Cy + 8.25 – 3 – 4.5 - = 0 Cy = 0.75 KN

BD and EA are zero force members, so

F3 = F8 = 0

Node C:

(0.9/1.5)F4 = 0.75 F4 = 1.25 KN (Tension)

F1 +(1.2/1.5) (1.25) = 0

F1 = - 1 KN = 1 KN (compression)

Node B :

F1 = F2 = 1 KN (compression)

Node D:

(0.9/1.5)F5 + 0.75 + 3 = 0

F5 = -6.25 KN = 6.25 KN (compression)

1 – F6 – 5 = 0

F6 = - 4 KN = 4KN (compression)

Node E:

F6 = F7 = 4 KN (compression)

Node F:

(0.9/1.5 )F9 = 4.5

F9 = 7.5 KN (Tension)

EX(7):

Determine the force in members GE, GC, and BC of the truss.

Indicate whether the members are in tension or compression.

Solution •Choose section a-a since it cuts through the three members

•Draw FBD of the entire truss

+∑Fx = 0; 400N - Ax = 0 Ax = 400N

∑MA = 0; -1200(8) - 400(3) + Dy(12) = 0 Dy = 900N

∑ +Fy = 0; Ay – 1200 + 900 =0 Ay = 300N

• Draw Free Body Diagram for the section portion

∑MG = 0; - 300(4) – 400(3) + FBC(3) = 0 FBC = 800N (Tension)

∑MC = 0; - 300(8) + FGE(3) = 0 FGE = 800N (Compression)

+ ∑Fy = 0; 300 - 35 FGC = 0 FGC = 500N (Tension)

EX(8):

Determine the force CF of the truss shown in the figure. Indicate whether the member is in tension or compression. Assume

each member is pin connected.

SolutionFree Body Diagram:

Section aa will be used since thissection will “expose” the internal force in member CF as “external”·on the free-body diagram of either the right or left portion of thetruss. It is first necessary, however, to determine the support reactionson either the left or right side.

The free-body diagram of the right portion of the truss, which is theeasiest to analyze, is shown in the following figure:

Equations of Equilibrium: We will apply the moment equationabout point 0 in order to eliminate the two unknowns F FG and FcD.The location of point 0 measured from E can be determined fromproportional triangles. i.e. 4/(4 + x) = 6/(8 + x ) . x = 4 m. Or,stated in another manner. The slope of member GF has a drop of 2 mto a horizontal distance of 4 m. Since FD is4 m. Fig. then fromD t0 O the distance must be 8 m.

An easy way to determine the moment of FCF about point O is to usethe principle of transmissibility and slide FCF to point C. and thenresolve FCF into its two rectangular components. We have

+ ∑MO = 0;

-FCF sin45 (12) + 3(8) – 4.5(4) = 0

FCF = 0.589 KN (compression)

EX(9):

Determine the force in member EB of the roof truss shown in the figure. Indicate whether the member is in tension or compression.

Solution