What does the principle of transmissibility state?

A force can be moved anywhere along its line of action without changing its external effect on a rigid body

A force can be moved to any parallel line with no change

A force can be moved anywhere on the body

Forces on a body are always transmitted equally to supports

When is the principle valid?

For rigid bodies and external effects only

For deformable bodies and internal stresses

How is the moment of the 300 kN force at C(2, 1.5) about O found?

x·Fy - y·Fx = 2(150) - 1.5(259.81) = -89.71

How is the line of action of R located relative to O?

x·R_y - y·R_x = M_O, giving x = -0.87 m (y = 0) or y = -0.61 m (x = 0)

R acts through O, since all forces are on the plate

At the corner nearest the largest force

2071 Chaitra · 3+8 marks

Principle of transmissibility; resultant on plate ABCD

Short answer

Principle of transmissibility: a force's external effect on a rigid body does not change if moved along its line of action. Resultant: 98.13 kN at 34.84° below the horizontal, passing 0.87 m left of O and 0.61 m below O.

Describe the principle of transmissibility. Determine the magnitude, direction and position with respect to center 'O' of the resultant of the forces acting on the rectangular plate ABCD as shown in figure below. [Figure: rectangular plate ABCD, DC = 4 m, BC = 3 m, centre O; 300 kN at C at 30°; 200 kN at D at 60°; 100 kN·m couple; 100 kN horizontal at A; 150 kN at B at 45°.]

Principle of transmissibility

The external effect of a force on a rigid body (its tendency to move or rotate the body) does not change if the force is moved to any other point on its line of action, with the same magnitude and direction. Only the line of action matters, not the point of application. For example, pushing a box at the back or pulling it with a rope at the front along the same line gives the same motion. It is valid only for rigid bodies and for external effects (not for internal stress or deformation).

Resultant of the forces on plate ABCD

Reading of the figure: A(−2, −1.5), B(2, −1.5), C(2, 1.5), D(−2, 1.5) with O at the centre (DC = 4 m, BC = 3 m). Forces:

300 kN at C, up and to the right at 30° to the horizontal.
200 kN at D, down and to the left, 60° from the vertical side DA (so 30° below the horizontal).
100 kN at A, horizontal, to the right.
150 kN at B, down and to the left at 45° to the vertical.
Couple 100 kN·m, clockwise.

Components

Force	Fx (kN)	Fy (kN)	Moment about O = x·Fy − y·Fx
300 at C(2, 1.5)	259.81	150.00	2(150) − 1.5(259.81) = −89.71
200 at D(−2, 1.5)	−173.21	−100.00	−2(−100) − 1.5(−173.21) = +459.81
100 at A(−2, −1.5)	100.00	0	0 − (−1.5)(100) = +150.00
150 at B(2, −1.5)	−106.07	−106.07	2(−106.07) − (−1.5)(−106.07) = −371.25
Couple			−100.00
Sum	80.54	−56.07	+48.87

Resultant

R = √(80.54² + 56.07²) = 98.13 kN θ = tan⁻¹(56.07/80.54) = 34.84° below the horizontal (towards the right).

M_O = +48.87 kN·m (counter-clockwise).

Position

Moment of R about O: x R_y − y R_x = M_O

Cuts the horizontal through O (y = 0) at x = 48.87/(−56.07) = −0.87 m (0.87 m to the left of O).
Cuts the vertical through O (x = 0) at y = −48.87/80.54 = −0.61 m (0.61 m below O).

Answer: R = 98.13 kN at 34.8° below the horizontal, passing 0.87 m to the left of O on the horizontal through O (or 0.61 m below O on the vertical through O).

✓ Answer checked by Claude Opus 5.5 (high effort)

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