Question

Determine FBC , the magnitude of the force in member BC, using the method of sections. Assume for your calculations that each member is in tension, and include in your response the sign of each force that you obtain by applying this assumption.

Answer 1

I've attached an image depicting this truss.

P1 = 971 N and P2 = 433N

Answer:

BC = 358.67 N

Thus, BC is in tension

Explanation:

Assumptions;

Truss members are mass-less

Clockwise is the positive moment

Directions to the right and up are positive directions

Now, let's find the reaction forces at A and E

To find vertical reaction E_y, let's find sum of moments about A and equate to zero.

We will asssume E_y is positive

Thus;

0 = P_1(a) - P_2(a) - E_y(3a)

Making E_y the subject, we have;

E_y = (P_1(a) - P_2(a))/(3a)

a will cancel out, thus;

E_y = (P_1 - P_2)/3

From the attached image, P1 = 971 N and P2 = 433N

Thus;

E_y = (971 - 433)/3

E_y = 179.33 N

Now, to find A_y, let's sum forces in the vertical direction to zero.

We'll assume A_y is positive

Thus;

0 = A_y + E_y - P_1

A_y = P_1 - E_y

A_y = 971 - 179.33

A_y = 791.67 N

To find A_x, let's sum forces in the horizontal direction to zero.

We'll assume A_x is positive.

Thus;

A_x - P_2 = 0

A_x = P_2

A_x = 433 N

A_x = 433 N

Now let's imagine a vertical cut through the three members GH, GC, and BC. We can eliminate the truss members and forces to the right of this cut and replace the cut stubs with equivalent assumed tensors which I will call GH, GC, and BC to represent the forces in each respective member.

Thus;

The angle α from vertical of member GC is; tanθ = (a/2)/a

tanθ = 1/2

θ = tan^(-1)(1/2)

θ = 26.57°

Let's now sum all forces in the vertical direction and equate to zero;

0 = A_y - P_1 - GC•cosθ

GC = (A_y - P_1) / cosθ

GC = (791.67 - 971)/cos26.57

GC = -200.5 N

Thus, GC is in compression

Let's now sum moments about G and equate to zero which will allow us to determine BC.

Thus;

0 = A_y(a) - A_x(a) - BC(a) + GC(0) + GH(0)

BC = A_y - A_x

BC = 791.67 - 433

BC = 358.67 N

Thus, BC is in tension

Let's sum forces in the horizontal direction and equate to zero.

0 = A_x + BC + GCsinθ + GH

-GH = A_x + BC + GCsinθ

-GH = 433 + 358.67 + (-200.5sin26.57)

-GH = 702

GH = -702 N

Thus, GH is in compression