Determine the members forces.

Question

Please help me solve

Answer 1

EXTERNAL FORCES

I am assuming that where no external force is indicated, the external force is 0.

Notation:

_pF_q denotes external force acting on node p directed towards node q.

We are given

_BF_C = 12

_BF_A = 25

_CF_F = 50

_DF_E = 25

We are to compute _AF_F, _AF_B, and _EF_D.

All forces are vectors; the notation _pF_q denotes the magnitude of the vector.

The direction of the vector is taken from the diagram.

If the solution makes a force negative, it means that it is in reality directed in the opposite direction.

We solve for the three forces by requiring that the net force acting on the

whole structure be 0. (Meaning that the structure is not flying away.)

And we require that the net torques around A and around E be 0.

(Meaning that the structure is not rotating.)

We will use the convention that clockwise torque is positive and counter-clockwise torque is negative.

We compute the net force, which is a vector, by computing its x- and y-components,

that is, its projection on the positive x-axis and the positive y-axis.

That will give us 2 equations.

The two requirements that the two net torques be 0 give us two extra equations.

So in total we have four equations and three unknowns ( _AF_F, _AF_B, and _EF_D).

It will turn out that the four equations are linearly dependent.

Horizontal component of net force:

_BF_C + _AF_F = 0

Vertical component of net force:

-_BF_A - _CF_F - _DF_E + _AF_B + _EF_D = 0

Net torque around A:

5_BF_C + 5_CF_F + 10_DF_E - 10_EF_D = 0

Net torque around E:

5_BF_C - 10_BF_A - 5_CF_F + 10_AF_B = 0

Substitute actual numbers:

12 + _AF_F = 0

-25 - 50 - 25 + _AF_B + _EF_D = 0

5×12 + 5×50 + 10×25 - 10_EF_D = 0

5×12 - 10×25 - 5×50 + 10_AF_B = 0

Solution:

_AF_F = -12

_EF_D = 56

_AF_B = 44

INTERNAL FORCES

Notation:

_pT_q denotes tension in beam connecting nodes p and q, and acting on node p.

As above, _qT_p denotes only the magnitude of the tension force.

As is common, all the tension forces are presumed to be directed towards the inside.

This is the reality for beams experiencing extension.

If our solution makes a tension force negative, it means that the beam experiences compression.

The structure contains 9 beams, which gives 18 tension variables to solve for.

For the beam between p and q we have two tension forces _pT_q and _qT_p.

However, by Newton's third law

_pT_q = _qT_p

which reduces the number of unknowns to 9.

For the structure to be stable the net force acting at each node must be 0.

As we did for the external forces, we can require that both and x- and y-components

of the tension forces add up to 0.

That gives us two equations per node, for a total of 12 equations.

As for the external forces, the resulting set of equations is linearly dependent.

We could write down all 12 equations and then solve the whole system.

It turns out that there is a simpler way:

There is always a node which has two or fewer unknown tension forces,

and those can be found using the two equations for that node.

Solving one node gives us values of forces at neighboring nodes, so this method can continue.

We are now going to following this procedure one node at a time.

For each node we will write the projection of the net force on the x-coordinate and on the y-coordinate.

Note that if T is the tension force in one the diagonal beams then its projection on both the x- and y-axis is T/√2.

The reason is that the rectangles AFCB and FEDC are actually squares (although not drawn that way).

node A:

Solve forces around node A because it has only two unknown tension forces -- _AT_B and _AT_F.

Horizontal components:

_AF_F + _AT_F = 0

Vertical components:

_AF_B + _AT_B = 0

Substitute actual numbers:

-12 + _AT_F = 0

44 + _AT_B = 0

Solution:

_AT_F = 12

_AT_B = -44

node B:

Solve forces around B because it has only two unknown tension forces -- _BT_F and _BT_C.

Horizontal components:

_BF_C + _BT_C + _BT_F/√2 = 0

Vertical components:

-_BF_A - _BT_A - _BT_F/√2 = 0

Substitute actual numbers:

12 + _BT_C + _BT_F/√2 = 0

-25 + 44 - _BT_F/√2 = 0

Solution:

_BT_F = 19√2

_BT_C = -31

node E:

Solve forces around E because it has only two unknown tension forces -- _ET_D and _ET_F.

Horizontal components:

-_ET_F = 0

Vertical components:

_EF_D + _ET_D = 0

Substitute actual numbers:

-_ET_F = 0

56 + _ET_D = 0

Solution:

_ET_F = 0

_ET_D = -56

node D:

Solve forces around D because it has only two unknown tension forces -- _DT_C and _DT_F.

Horizontal components:

-_DT_C - _DT_F/√2 = 0

Vertical components:

-_DF_E - _DT_E - _DT_F/√2 = 0

Substitute actual numbers:

-_DT_C - _DT_F/√2 = 0

-25 + 56 - _DT_F/√2 = 0

Solution:

_DT_F = 31√2

_DT_C = -31

node C:

Solve forces around node C because it has only one unknown tension forces -- _CT_F.

We use the equation for horizontal forces only as a consistency check.

Horizontal components:

-_CT_B + _CT_D = 0

Vertical components:

-_CF_F - _CT_F = 0

Substitute actual numbers:

-31 + 31 = 0

-50 - _CT_F = 0

Solution:

_CT_F = -50

node F:

We have already found all the tension forces.

Use the equations for node F only as a consistency check.

Horizontal components:

-_FT_A + _FT_E - _FT_B/√2 + _FT_D/√2 = 0

Vertical components:

_FT_C + _FT_B/√2 + _FT_D/√2 = 0

Substitute actual numbers:

-12 + 0 - 19 + 31 = 0

-50 + 19 + 31 = 0