_{1}

^{*}

The development of a theoretical model to predict the four equilibrium forces of reaction on a simple ladder of non-adjustable length leaning against a wall has long remained an unresolved matter. The difficulty is that the problem is statically indeterminate and therefore requires complementary information to obtain a unique solution. This paper reports 1) a comprehensive theoretical analysis of the three fundamental models based on treating the ladder as a single Euler-Bernoulli beam, and 2) a detailed experimental investigation of the forces of reaction as a function of applied load and location of load. In contrast to previous untested proposals that the solution to the ladder problem lay in the axial constraint on compression or the transverse constraint on flexure, the experimental outcome of the present work showed unambiguously that 1) the ladder could be modeled the best by a pinned support at the base (on the ground) and a roller support at the top (at the wall), and 2) the only complementary relation needed to resolve the static indeterminacy is the force of friction at the wall. Measurements were also made on the impact loading of a ladder by rapid ascent and descent of a climber. The results obtained were consistent with a simple dynamical model of the ladder as a linear elastic medium subject to a pulse perturbation. The solution to the ladder problem herein presented provides a basis for theoretical extension to other types of ladders. Of particular importance, given that accidents involving ladders in the workplace comprise a significant fraction of all industrial accidents, the theoretical relations reported here can help determine whether a collapsed structure, against which a ladder was applied, met regulatory safety limits or not.

Although determination of the reaction forces on a beam is a standard part of the mechanics of continuous media, the exact conditions under which an actual physical structure comprised of beams is stressed are not always apparent. Models based on different assumptions can predict hugely different reaction forces, and careful experimentation is required to determine which model may most accurately characterize a structure. One of the most important mechanical structures for which reaction forces need to be determined is that of a ladder. Although ladders have been in use since antiquity, to date no definitive model of the reaction forces on a ladder has been experimentally tested and confirmed. This paper reports what the author believes to be the most comprehensive theoretical analysis and definitive experimental test of the reaction forces on a fixed ladder.

A fixed ladder is defined by the American Ladder Institute as a non-self-supporting ladder of fixed (i.e. non-adjustable) length [

w = W / L (1)

that acts vertically downward at the center of mass of any isolated segment of the beam. A climber is modeled as a point load P acting vertically downward at a distance β L ( 1 ≥ β ≥ 0 ) along the beam from the contact point at the ground, which is taken to be the origin.

As shown in panel A of the figure,there are four reaction forces:1) normal force at the ground ( R 1 ) , 2) parallel force at the ground ( R 2 ) , 3) normal force at the wall ( R 3 ) , and 4) parallel force at the wall ( R 4 ) . The prediction of these four forces constitutes what is called the ladder problem. Before discussing the theoretical solutions and experimental test of the ladder problem, it is worthwhile to consider why the problem is a significant one. This significance is both conceptual and practical.

First, conceptually, the solution to any model serving as an archetype for a class of real-world problems within some branch of physics is significant by virtue of its fundamentality. The single-beam ladder is an archetype within the mechanics of continuous media. Moreover, understanding the forces on a fixed

ladder would provide useful modeling guidelines for predicting the reaction forces on more complex ladder types, such as extension ladders, step ladders, trestle ladders, and others. Second, from a practical standpoint, the ability to predict the forces on ladders has direct impact on matters relating to safety in the use of ladders and to legal issues that arise when use of a ladder in the workplace results in injury or death. For example, with a valid fixed ladder model, the analyst can determine the acceptable range of inclination angles to avoid slippage, or determine whether the net force exerted by a ladder on a wall or railing that collapsed was within or exceeded regulatory limits [

One might think that prediction of the reaction forces on a structure as simple as a single-beam ladder is a relatively straightforward matter. Indeed, the problem is trivially solvable if the parallel reaction at the wall is ignored, which is the case in countless physics textbooks (see, for example, [

When account is taken of the parallel reaction at the wall, the problem becomes statically indeterminate, and therefore more complicated, since there are now 4 unknowns and 3 equations. One proposed solution assumed that the ladder undergoes an axial deformation which is constrained by the supports [

Experimentally, there have been many investigations, of which some representative examples are [

This paper reports the theoretical solution of three basic models by which to analyze the reaction forces of a fixed ladder represented as a single Euler-Bernoulli (E-B) beam. The characteristics of an E-B beam pertinent to the present study are as follows [

1) the length of the beam is much greater than the linear dimension of the cross-section;

2) the long axis of the beam lies within the neutral surface;

3) the cross section of the beam remains plane and perpendicular to the long axis during deflection;

4) deformation of the cross section within its own plane is neglected;

5) the beam is a linear elastic material subject to Hooke’s law; i.e. normal stress within a cross section varies linearly with perpendicular distance from the neutral axis.

The neutral surface of a beam is the interface that separates the fibers under compression from the fibers under tension when the beam is deflected transversely. The neutral axis within any cross section is the line of intersection of the neutral surface with the cross section.

The three basic ladder models analyzed in this paper are distinguished by the nature of the supports at the points of contact with the surfaces of the ground and wall. In the mechanics of beams, the three common types of supports are 1) roller, 2) pin, and 3) fixed [

If a right-handed Cartesian coordinate system ( x , y , z ) is chosen such that an initially horizontal beam lies along the x axis, and a load is applied that can deflect the beam vertically downward along the negative y axis, then the boundary and continuity conditions at the three kinds of supports can be summarized as follows

Roller : Δ y = 0 , d y / d x ≠ 0 , M z = 0 Pin : Δ y = 0 , d y / d x ≠ 0 , M z = 0 Fixed : Δ y = 0 , d y / d x = 0 , M z ≠ 0 (2)

where the symbol Δ signifies displacement, and M z is an internal bending moment (torque) about the z axis. The salient features of relations (2) are that a) the displacement and slope of the deflected beam are zero at a fixed support but the support creates an internal bending moment, whereas b) at roller and pin supports there is no internal moment, and the slope of the bending curve there need not be zero.

The models analyzed in this paper are based on the supports described by Equation (2).

・ In Model 1, the ground support of the single-beam ladder is fixed and the wall support is equivalent to a pin. Under a vertical load due to the distributed weight of the ladder and the point weight of a hypothetical climber, the ladder is subject to both axial and transverse stresses, but is constrained from compression, elongation, or rotation at the supports. This is a case of static indeterminateness in two variables: one axial and one transverse. Although friction is presumably the source of the constraints, no specific assumptions are made in regard to the mathematical form of the frictional force.

・ In Model 2, the ground support of the single-beam ladder is fixed, and the wall support is equivalent to a roller. The axial component of the load can compress the ladder, and thereby remove the static indeterminateness along the length. The reaction at the wall is assumed to be governed by the phenomenological relation for friction taken to be proportional to the normal force at the wall [

・ In Model 3, the ground support of the single-beam ladder is equivalent to a pin and the wall support is equivalent to a roller. The ladder is free to compress under an axial force and flex about the ground support, thereby relieving all static constraints. The reaction at the wall is assumed governed by friction as in Model 2. This is a case of static indeterminateness in one variable in which the equations of static equilibrium, complemented by the frictional force law, suffice to determine a unique solution.

The three models are schematically illustrated in

In Section 2 the reaction forces on a ladder characterized as a single E-B beam are calculated for each model and examined as a function of load, location of load along the ladder, and angle of inclination of the ladder with respect to the ground. In the case of Models 1 and 3, the full bending curve is calculated within the Euler-Bernoulli approximation (neglect of the square of the slope d y / d x ), and the magnitudes of transverse and axial deformations are compared. (The bending curve of Models 2 and 1 are the same).

In Section 3 are reported experiments using horizontal and vertical force platforms to measure the reactions on a fixed metal ladder inclined against a wall. It is noted here briefly that the model that was found to account for the observed results was Model 3.

In Section 4 the impact loading on a ladder by a climber is examined theoretically and experimentally.

In Section 5 are reported measurements of the reactions on a wood beam inclined obliquely against a wall. The intent of the experiment was to ascertain whether the same theoretical model that best applied to a ladder modeled as a beam applied as well to an actual single beam.

In Section 6 conclusions drawn from this research are summarized.

Panel A of

A 1 = R 1 sin θ + R 2 cos θ B 1 = R 1 cos θ − R 2 sin θ A 2 = − R 3 cos θ + R 4 sin θ B 2 = R 3 sin θ + R 4 cos θ (3)

with inverse relations

R 1 = B 1 cos θ + A 1 sin θ R 2 = − B 1 sin θ + A 2 cos θ R 3 = − B 2 sin θ − A 2 cos θ R 4 = B 2 cos θ + A 2 sin θ (4)

The equations of static equilibrium then take either of the forms

(I) R 1 + R 4 = W + P R 2 − R 3 = 0 (II) B 1 + B 2 = W n + P n A 1 + A 2 = W a + P a (5)

in which the normal and axial components of the ladder weight and a point load representative of a climber are respectively

W n = W cos θ W a = W sin θ P n = P cos θ P a = P sin θ . (6)

The sets of equations in (5) do not contain an equation for the vanishing of a rotational moment about a designated point at rest. In the absence of information regarding the supports of the ladder at the ground, it is not possible to write a torque equation based on static equilibrium alone. This is a consequence of the fact that the system may contain an initially unknown internal moment created by the ground support in order to maintain static equilibrium.

Since there are four unknown reactions in Equation (5) and fewer than four equations of static equilibrium, the set of equations is statically indeterminate and must be supplemented by what are termed complementary conditions. Under the circumstances stated above, these conditions can be applied to the axial and normal directions independently. The complementary conditions take different forms depending on the specific model, as discussed in the following subsections.

With respect to the schematic diagram in panel B of

U = U a + U b (7)

is the sum of the axial strain energy

U a = 1 2 E A ∫ 0 L F a ( x ) 2 d x (8)

due to the compression or tension by force F a ( x ) and the energy of flexure (bending)

U b = 1 2 E I z ∫ 0 L M z ( x ) 2 d x (9)

due to the moment M z ( x ) about the z axis (see Ref. [

A = w h (10)

I z = 1 12 w h 3 . (11)

Because the overall load comprises both distributed and point contributions, the force F a ( x ) and moment M z ( x ) are piecewise continuous functions. These functions can be represented economically by use of the Heaviside unit step function [

H ( x − x 0 ) = { 0 x < x 0 1 x ≥ x 0 . (12)

Mathematically, the Heaviside function is usually defined by pure inequalities in both partitions of the real axis and left undefined at the point where the argument is zero. In the mechanics of continuous media, however, the point load occurs at the partition between segments and therefore it is more useful to define the Heaviside function as unity there. One can then express the axial force and bending moment in the following ways

F a ( x ) = A 2 − W a L ( L − x ) − P a H ( β L − x ) (13)

M z ( x ) = B 2 ( L − x ) − W n 2 L ( L − x ) 2 − P n ( β L − x ) H ( β L − x ) . (14)

From the complementary energy theorem (Castigliano’s theorem) of continuum mechanics (see [

δ i = ∂ U ∂ F i . (15)

Since a physical condition of Model 1 is that the right endpoint of the beam in panel B of

δ L ( a ) = ∂ U ∂ A 2 = 0 (16)

and

δ L ( b ) = ∂ U ∂ B 2 = 0 (17)

where superscript a signifies axial and superscript b signifies bending. Equations (16) and (17) are the complementary relations required to supplement the set of relations (5) and resolve the static indeterminacy of Model 1.

Substitution of relations (13) and (14) respectively into strain energies (8) and (9), taking derivatives (16) and (17), and setting the resulting integrals to zero lead to expressions for the axial reaction A 2 and transverse reaction B 2 . The reactions A 1 and B 1 are then obtained from Equation (5) of static equilibrium. The resulting set of four axial and transverse reactions are found to be

A 1 = 1 2 W sin θ + ( 1 − β ) P sin θ A 2 = 1 2 W sin θ + β P sin θ B 1 = 5 8 W cos θ + ( 1 − 3 2 β 2 + 1 2 β 3 ) P cos θ B 2 = 3 8 W cos θ + ( 3 2 β 2 − 1 2 β 3 ) P cos θ (18)

from which follow by transformation (4) the reactions normal and parallel with respect to the ground and wall

R 1 = 1 2 W ( 1 + 1 4 cos 2 θ ) + P [ 1 − ( 3 2 β 2 − 1 2 β 3 ) cos 2 θ − β sin 2 θ ] R 2 = R 3 = [ 1 16 W + ( 1 2 β − 3 4 β 2 + 1 4 β 4 ) P ] sin 2 θ R 4 = 1 2 W ( 1 − 1 4 cos 2 θ ) + P [ ( 3 2 β 2 − 1 2 β 3 ) cos 2 θ + β sin 2 θ ] . (19)

The directions of R 2 and R 3 are shown in

A complementary perspective is given in

fixed weight) influences the normal reaction at the wall strongly and the normal reaction at the ground relatively weakly as the inclination of the ladder is changed.

Applying Equation (15) to the total strain energy U, one can determine the deformation at any point in the elastic curve of the single-beam ladder and thereby compare the magnitudes of the axial compression δ ( a ) and transverse

(bending) deflection δ ( b ) . Taking β = 1 2 for illustration yields the deformations

δ 1 2 L ( a ) = ∂ U ∂ P a | β = 1 2 = L 2 A E ( W + P ) sin θ (20)

δ 1 2 L ( b ) = ∂ U ∂ P n | β = 1 2 = L 3 768 E I z ( 4 W + 7 P ) cos θ . (21)

Given typical parameters of beam length L = 2.5 m, width w = 2 cm, height h = 5 cm, inclination angle θ = 45 ∘ , ladder weight W = 25 lbs (111.2 N), climber weight P = 150 lbs (667.2 N), and elastic modulus of aluminum

E = 69 GPa, Equations (20) and (21) lead to deformations δ 1 2 L ( a ) ≈ 10 µm, δ 1 2 L ( b ) ≈ 5.1 mm. The deformation due to bending is approximately 500 times the

deformation due to axial compression. Therefore, any model of a ladder subject to ordinary loads that takes account of axial deformation cannot neglect deformation due to flexure.

As a final consideration, it is useful to derive the deflection curve of Model 1 as a check on the calculated deformations and to ascertain that the reactions of a single-beam ladder derived from Model 1 correctly fulfill the complementary conditions. The exact flexure formula for an E-B beam relates the bending moment to the radius of curvature and takes the form of a nonlinear second-order differential equation (see [

d 2 y / d x 2 [ 1 + ( d y / d x ) 2 ] 3 / 2 = 1 E I z M z ( x ) (22)

whose solution is called the elastica [

y ( 0 ) = y ( L ) = 0 y ( x ) i s c o n t i n u o u s a t x = β L d y / d x iscontinuousat x = β L (23)

to yield the piecewise continuous function

y ( x ) = y 1 ( x ) H ( β L − x ) + y 2 ( x ) H ( x − β L ) (24)

with

y 1 ( x ) Model1 = 1 E I z [ − L ( 1 16 W n + ( 1 2 β − 3 4 β 2 + 1 4 β 3 ) P n ) x 2 + ( 5 48 W n + 1 6 ( 1 − 3 2 β 2 + 1 2 β 3 ) P n ) x 3 − W n 24 L x 4 ] y 2 ( x ) Model1 = 1 E I z [ 1 6 β 3 L 3 P n − 1 2 ( β 2 L 2 P n ) x + L ( − 1 16 W n + ( 3 4 β 2 − 1 4 β 3 ) P n ) x 2 + ( 5 48 W n − 1 6 ( 3 2 β 2 − 1 2 β 3 ) P n ) x 3 − W n 24 L x 4 ] . (25)

ladder x = L . The deflection δ 1 2 L ( b ) ∼ 5.1 mm occurs at the mid-position β = 1 2 ,

which is the result predicted by Equation (21) obtained by Castigliano’s theorem. The deflection curve is not symmetric for any location of the load, even at mid-position, because the slope is required to be 0 at the ground support (left end) but not at the wall support (right end).

Calculation of deflections (20), (21) and the elastica curve (24) involved

integration of expressions of the form x n H ( x − x 0 ) for non-negative integer n. The integration of Heaviside functions is summarized in the Appendix.

With respect to the schematic diagram in panel B of

R 4 = μ R 3 (26)

withcoefficient of friction μ to obtain

B 2 = R 3 sin θ + R 4 cos θ = R 3 ( sin θ + μ cos θ ) , (27)

fromwhich follows

R 3 = R 2 = R 4 μ = 3 8 W + ( 3 2 β 2 − 1 2 β 3 ) P tan θ + μ (28)

andtherefore from Equation (5)

R 1 = W + P − R 4 = W + P − μ R 3 = W + P − μ [ 3 8 W + ( 3 2 β 2 − 1 2 β 3 ) P tan θ + μ ] . (29)

The axial forces A 1 and A 2 are given by the transformation (3)

A 1 = ( W + P ) sin θ + ( 3 8 W + ( 3 2 β 2 − 1 2 β 3 ) P cos θ ) [ 1 − μ tan θ tan θ + μ ] A 2 = − ( 3 8 W + ( 3 2 β 2 − 1 2 β 3 ) P cos θ ) [ 1 − μ tan θ tan θ + μ ] (30)

ratherthan by a complementary condition constraining axial compression.

Since the reaction forces normal to the beam ( B 1 , B 2 ) are the same as for Model 1, the bending deflection curve, which does not depend on the axial reactions, is also the same.

With respect to the schematic diagram in panel B of

pinned support at the left end and a roller support at the right end. Because there is no unknown internal moment at pinned and roller supports, the set of equations (5) can be supplemented by an equation for the vanishing of the external moment M z about any fixed point. For example, if the pivot is taken to be the point of contact with the ground, then the vanishing of M z ( 0 ) can be expressed by either of the equations

R 3 sin θ + R 4 cos θ = 1 2 W cos θ + β P cos θ (31)

or

B 2 = 1 2 W n + β P n . (32)

Together with the complementary relation (26) for force of friction and the other equations of static equilibrium, one obtains

R 3 = R 2 = R 4 μ = 1 2 W + β P tan θ + μ (33)

B 1 = 1 2 W n + ( 1 − β ) P n (34)

and

R 1 = W + P − μ R 3 = W ( tan θ + 1 2 μ ) + P ( tan θ + ( 1 − β ) μ ) tan θ + μ . (35)

From the set of transformations (3) follow the axial reactions

A 1 = ( W + P ) sin θ + ( 3 8 W + ( 3 2 β 2 − 1 2 β 3 ) P cos θ ) [ 1 − μ tan θ tan θ + μ ] A 2 = − ( 3 8 W + ( 3 2 β 2 − 1 2 β 3 ) P cos θ ) [ 1 − μ tan θ tan θ + μ ] (36)

The bending curve (elastica) of Model 3 takes a different shape than that of Models 1 and 2 because of the difference in the nature of the support at the ground. Solving the Euler-Bernoulli approximation to the flexure equation (22)

d 2 y / d x 2 = M z ( x ) / E I z (37)

with moment (see panel B of

M z ( x ) = B 1 x − W n 2 L x 2 − P n ( x − β L ) H ( x − β L ) (38)

and boundary conditions (23) leads to the piecewise continuous function (24) with components

y 1 ( x ) Model3 = 1 E I z [ − L 2 ( 1 24 W n + ( 1 3 β − 1 2 β 2 + 1 6 β 3 ) P n ) x + ( 1 12 W n + 1 6 ( 1 − β ) P n ) x 3 − W n 24 L x 4 ] y 2 ( x ) Model3 = 1 E I z [ β 3 L 3 6 P n − L 2 ( 1 24 W n + ( 1 3 β + 1 6 β 3 ) P n ) x + ( 1 2 β L P n ) x 2 + ( 1 12 W n − 1 6 β P n ) x 3 − W n 24 L x 4 ] (39)

The maximum deflection δ 1 2 L ( b ) ∼ 11.8 mm occurs at the mid-position

β = 1 / 2 . This is the same result one would get from applying Castigliano’s theorem, as implemented in Section 2.1, which leads to the deformation

δ 1 2 L ( b ) = L 3 384 E I z ( 5 W + 8 P ) cos θ . (40)

The deflection produced by the axial forces of Equation (36) is given by

δ β L ( a ) = L A E ( A 2 − ( 1 2 W + β P ) sin θ ) ( 1 − μ tan θ tan θ + μ ) . (41)

For the load at mid-position and the other parameters of

To determine which, if any, of the three basic models describes most closely the reaction forces on a fixed ladder, an experiment was performed to measure forces R 1 , R 2 , and R 3 . The experiment was performed as follows. A ladder of length 244 cm and weight 18.9 lbs (84.07 N) was placed against a vertical wall at an angle of 68˚ to the ground. A two-axis force platform (Pasco Model PS-2142) was situated horizontally on the ground beneath each of the two rails of the ladder. The top of the ladder rested on a third force platform attached vertically to the wall. Each force platform, used in conjunction with a computer and the appropriate interface, was designed to measure the applied force normal to its surface (up to a maximum of 4400 N) and along an axis parallel to its surface (up to 1100 N).

The ladder comprised seven rungs located at fractional lengths along the rails of

β = ( 32 244 , 62 244 , 92 244 , 122 244 , 152 244 , 182 244 , 212 244 ) ≈ ( 0.131 , 0.254 , 0.377 , 0.500 , 0.623 , 0.746 , 0.869 )

of which the first six rungs were used in the experiment. External loads were applied to the rungs in the form of lead bricks, each brick weighing 25 lbs (111.21 N). Measurements of the normal and parallel forces were made for discrete loads of 1, 2, 3, or 4 bricks placed on each rung in progressive order from rung 1 to rung 6. The sum of the forces normal to the horizontal surface of force platforms 1 and 2 (supporting the rails) constituted reaction R 1 . The sum of the forces parallel to the surface of platforms 1 and 2 constituted reaction R 2 . The force normal to the vertical surface of force platform 3 (against which the top of the ladder rested) constituted reaction R 3 . Although the initial intention was to measure reaction R 4 as well, this measurement could not be made because the force platforms did not measure shear forces accurately when the platform was oriented vertically^{1}. The omitted reaction, however, was not needed to establish the conclusions drawn from this experiment.

In order to apply the theoretical relations of Models 2 and 3, it is necessary to know the coefficient of friction μ of the ladder with the wall―or, more accurately, with the surface of the force platform attached to the wall. This coefficient cannot be calculated theoretically, but must be obtained empirically from the data. The method employed here was trial and inspection; i.e. to assume a value for μ , calculate and plot the resulting reactions, and adjust as necessary the value of μ until a satisfactory match with data was achieved, provided that the model itself is a valid representation of the underlying mechanical processes. The entire procedure can be executed in a matter of seconds with a fast computer and available mathematical software such as Maple or Mathcad. It is to be emphasized that the totality of the 72 measurements (3 reactions × 4 loads × 6 load locations) tightly constrained the value of μ . For Model 2 or 3 to be valid it is necessary that a single value of μ suffice to account for all the data. If different statistically significant values of μ for the various reactions are needed to match the data, or if no value of μ matches the data, then Model 2 or 3 would not describe the static equilibrium of a ladder against a wall.

θ = 68.0 ∘ | LOAD (lbs) | 25 | 50 | 75 | 100 |
---|---|---|---|---|---|

REACTIONS (N) | RUNG | ||||

Ground Normal R 1 | |||||

1 | 347.81 | 303.74 | 412.54 | 526.04 | |

2 | 188.54 | 302.98 | 413.19 | 526.67 | |

3 | 193.31 | 304.92 | 409.82 | 522.06 | |

4 | 189.95 | 303.37 | 413.52 | 520.52 | |

5 | 187.31 | 300.74 | 409.77 | 516.84 | |

6 | 188.50 | 298.47 | 403.50 | 508.11 | |

Ground Parallel R 2 | |||||

1 | 20.19 | 23.87 | 31.56 | 37.00 | |

2 | 22.92 | 35.01 | 47.29 | 58.86 | |

3 | 32.47 | 48.67 | 63.63 | 80.60 | |

4 | 37.15 | 60.90 | 86.58 | 106.12 | |

5 | 37.50 | 67.05 | 95.88 | 121.87 | |

6 | 47.62 | 81.22 | 113.40 | 145.17 | |

Wall Normal R 3 | |||||

1 | 10.01 | 14.09 | 17.70 | 21.40 | |

2 | 16.45 | 28.49 | 36.74 | 46.61 | |

3 | 24.14 | 39.44 | 53.67 | 69.85 | |

4 | 28.82 | 51.80 | 74.75 | 95.94 | |

5 | 32.76 | 60.15 | 87.46 | 112.59 | |

6 | 40.07 | 73.52 | 104.20 | 138.55 |

From examination of Figures 11-13 one can conclude the following:

・ Model 1 does not account for any of the measured reactions.

・ Models 2 and 3 account more or less equally well for the normal reaction R 1 at the ground support (

・ Model 2 does not account for the measurements of the parallel reaction R 2 at the ground (

・ Model 3 accounts satisfactorily for all three reaction forces for μ ≈ 0.25 estimated from the data by trial and inspection.

To test the consistency of Model 3, the reaction parallel to the wall R 4 was obtained from the data by the two independent relations

( a ) R 4 = μ R 3 ( b ) R 1 + R 4 = W + P (42)

where relation (a) is one of the defining conditions of the model, and relation (b) is required by Newton’s 2nd law applied to static equilibrium. Panel A of

From the foregoing comprehensive experimental test, one can conclude that, of the various mechanical conditions presumed to describe the static equilibrium of a ladder against a wall, those upon which Model 3 is based best represent the actual physical situation. It is useful to summarize explicitly the nature of these conditions.

・ The elimination of Model 1 indicates that the length of the ladder is not constrained by axial forces at the ground or wall. Recall that it was this constraint that provided a complementary relation for resolving static indeterminateness in the axial direction. However, the reported experiment is consistent with no unrelieved axial strain. This means that the ladder, in response to its own weight and/or to the applied load, has relaxed to its (very slightly) compressed length in the course of reaching static equilibrium.

・ The elimination of Model 2 (and Model 1) likewise indicates that the bending of the ladder does not engender a torque at the ground support. Recall that it was this initially unknown moment that necessitated a complementary relation to resolve the static indeterminateness in the transverse direction. The reported experiment is consistent with no internal rotational moment at the base. This means that the ladder, initially presumed to be straight, has relaxed (very slightly) to the shape of an elastic deformation curve like those shown in

・ The satisfactory accord between Model 3 and the experimental results tends to confirm that, for a ladder stationary, but not clamped, at its base, it is the friction at the wall that provides the complementary relation needed to resolve static indeterminateness so that all four forces of reaction can be calculated. The theoretical expressions for the reactions do not depend on the elastic modulus of the ladder. They do depend, however, on the coefficient of friction of the ladder at the wall. The experimental results are consistent with an empirical relation whereby the friction at a surface is proportional to the normal force on that surface.

The theoretical models tested in this paper each embodied one or more distinct physical attributes conceivably responsible for the forces of reaction on a fixed ladder modeled by a single beam. In this regard, the models were never intended to provide an exact theory of the stresses and strains within an actual ladder. To achieve that would require a detailed mechanical model, outside the scope of this paper, taking account of the exact shape, weight distribution, and area moments of inertia of the components of the ladder, including rungs. This caveat must be borne in mind when comparing theory and experiment in Figures 11-14. The dashed lines signifying Model 3 in the figures do not pass through all the experimental points, although they are sufficiently close for the purposes of distinguishing among the models. Small adjustments in the value of the coefficient of friction μ could achieve better agreement in one reaction or another, but there is little reason to make such refinements. Nor is there any reason to attempt to obtain an optimum value of μ by a statistical fitting process such as the method of maximum likelihood, or least square’s analysis, or by Bayes’ theorem [

The great practical utility of Model 3, now that it is validated by experiment, is that it provides a simple, but satisfactory, way to estimate the forces exerted by ladders on structures as a function of the climber’s weight and position of ascent and the ladder’s angle of inclination. Moreover, it is readily generalizable in special situations of practical interest that may arise in engineering design or matters of litigation, such as cases where the ground is flat but sloped, or the wall is flat but not vertical, or the ladder inclines against a railing, not a wall, and overhangs it.

Although the primary focus of this paper has been on the static equilibrium of a ladder against a wall, the manner in which ladders are used in practice makes it necessary to discuss, albeit briefly, the question of impact loads. The experimental data of

One sees, however, significant fluctuations about the equilibrium values as the climber transitions from one rung to the next. In particular, on the ascent, the reactions at the wall drop suddenly as the climber lifts his foot from a rung, and then rises suddenly as he places his foot on the next higher rung. On the descent, the reverse occurs for the normal reaction at the wall: the reaction increases dramatically as the climber raises his foot to step on the rung below.

The conveyance of momentum to a structure over a time interval is referred to as impact or impulse loading [

matter in the safe use of a ladder is to determine how large a force the impulsive loading by a climber creates in comparison to the forces of static equilibrium.

One way to do this is to model the ladder as linear elastic structure governed by Hooke’s law

P ( t ) = − k y ( t ) (43)

in which k is the force constant and y ( t ) is the deformation under the load P ( t ) . It is well known from the physics of oscillators [

k = M ω 2 (44)

where M is the mass of the object. Newton’s second law, in the absence of dissipation of mechanical energy, then takes the form

d 2 y d t 2 + ω 2 y = P ( t ) M (45)

for which the general solution, derivable by means of Green’s functions [

y ( t ) = 1 M ω ∫ 0 t sin ω ( t − t ′ ) P ( t ′ ) d t ′ (46)

with initial conditions

y ( 0 ) = d y d t | t = 0 = 0 . (47)

The impulse load P ( t ) can be modeled as a pulse of amplitude P 0 and width τ by the difference of two Heaviside functions

P ( t ) = P 0 [ H ( t ) − H ( t − τ ) ] . (48)

Substitution of Equation (48) into (46) leads to the solution

y ( t ) = P 0 M ω 2 [ cos ω ( t − τ ) − cos ω t ] (49)

from which one obtains by standard calculus the maximum deflection

y max = P 0 M ω 2 2 ( 1 − cos ω τ ) (50)

and therefore from Equations (43), (44) and (50) the maximum force of impact

P max = | M ω 2 y max | ω τ = π = 2 P 0 . (51)

P 0 is interpretable as the time-averaged load or, equivalently, the load at static equilibrium―i.e. the weight of the climber in the context of a ladder.

To the extent that the conditions of the pulse model realistically apply to a ladder, Equation (51) indicates that a climber can generate a maximum impact load that is twice his weight.

particular interest is the region within the black rectangle, which corresponds to the transition of the climber from rung 3 to rung 4 within a time interval Δ t of about 0.2 s. Since the vertical separation between the rungs is Δ s = 30 cm, the vertical acceleration a of the climber can be estimated from the kinematics of uniform acceleration [

a = 2 Δ s ( Δ t ) 2 ≈ 2 × 0.3 m ( 0.2 s ) 2 = 15 m ⋅ s − 2 (52)

or, relative to the acceleration of gravity g = 9.81 m ⋅ s − 2 ,

a g = P climber P 0 ≈ 15 10 = 1.5 . (53)

Equation (53) is consistent with Equation (51) of the dynamical pulse model. Indeed, in a more comprehensive model, the dissipation of mechanical energy must be taken into account and would lead to a ratio ( P max / P 0 ) < 2 . Nevertheless, the import of Equation (51) is that the numerical difference between impact forces and forces at static equilibrium can be consequential for those who use ladders in their work or who are responsible for setting safety guidelines and regulations with respect to ladders.

The single-beam models developed in Section 2 to describe the reaction forces on a ladder inclined against a wall should, in principle, apply to an actual single beam inclined against a wall. The author is unaware of any previously reported experiment to measure these forces and determine which, if any, of the theoretical models best accounts for these forces. This section reports the results of such a test.

Measurements using one horizontal force platform at the base and one vertically mounted force platform at the top were made on a wood beam of mass 6.92 kg, length L = 2.04 m, width w = 7.5 cm, height h = 8.5 cm (symbols correspond to panel C of

β = ( 42 204 , 72 204 , 102 204 , 132 204 , 162 204 , 192 204 ) ≈ ( 0.206 , 0.353 , 0.500 , 0.647 , 0.794 )

of which the first five eyelets were used in the experiment. Total loads (apart from the weight of the beam) ranged from about 4.4 N to 102.4 N.

This experiment lends strong support to the conclusion that the reaction forces on a single beam are described by the same theoretical model that describes the reaction forces on a fixed ladder, and that, as long as the base of either does not slip, the complementary relation needed to resolve the static indeterminateness of the problem is the force of friction at the wall.

θ = 57.7 ∘ | LOAD (N) | 43.55 | 63.15 | 82.75 | 102.35 |
---|---|---|---|---|---|

REACTIONS (N) | EYELET | ||||

Ground Normal R 1 | |||||

1 | 100.3 | 118.8 | 137.7 | 156.6 | |

2 | 99.6 | 118.6 | 137.3 | 154.6 | |

3 | 100.0 | 119.2 | 134.1 | 150.9 | |

4 | 98.0 | 115.8 | 131.8 | 149.0 | |

5 | 99.0 | 113.8 | 131.9 | 148.3 | |

Ground Parallel R 2 | |||||

1 | 21.8 | 23.1 | 15.9 | 29.5 | |

2 | 25.8 | 31.0 | 35.3 | 38.5 | |

3 | 31.9 | 36.8 | 46.9 | 49.4 | |

4 | 34.6 | 40.4 | 45.6 | 53.1 | |

5 | 36.2 | 44.3 | 53.6 | 62.6 | |

Wall Normal R 3 | |||||

1 | 22.7 | 25.0 | 28.2 | 31.1 | |

2 | 27.5 | 32.8 | 38.0 | 42.1 | |

3 | 33.3 | 40.1 | 45.7 | 51.5 | |

4 | 36.7 | 45.8 | 53.5 | 62.2 | |

5 | 42.4 | 44.3 | 63.0 | 73.0 | |

Wall Parallel R 4 | |||||

1 | 9.0 | 9.7 | 10.6 | 10.9 | |

2 | 9.2 | 10.3 | 10.7 | 13.0 | |

3 | 9.1 | 9.7 | 13.5 | 16.2 | |

4 | 10.7 | 12.8 | 16.1 | 18.4 | |

5 | 10.4 | 14.3 | 15.9 | 18.7 |

The question of how to calculate the four forces of reaction on a ladder in static equilibrium against a wall has been an unresolved issue for many years. The core of the difficulty is that the problem is statically indeterminate and requires complementary relations beyond the equations of static equilibrium. Three fundamental types of models, differing in the assumptions with regard to supports at the ground and wall, were examined. The comprehensive theoretical analysis and experimental measurements of this paper showed unambiguously that the forces of reaction were best accounted for by the model employing a single complementary relation characterizing friction at the wall.

The force law for friction that resolved the static indeterminateness in the third model is a single-parameter empirical relation. This parameter, the coefficient

of friction μ , was estimated from, and tightly constrained by, the experimental data. A critical test of the third model was, in fact, to show that a single value of this coefficient sufficed to yield the normal and parallel reactions at the ground and at the wall for all values and locations of the loads.

As part of the analyses reported here, the axial and transverse deformations of the single-beam ladder were calculated by means of Castigliano’s theorem, and then confirmed by a full calculation of the deflection curve (elastica) of the beam. The analyses demonstrated that the deformation due to flexure is about two to three orders of magnitude larger than the deformation due to compression. Thus, any model that takes into account only axial compression, but neglects flexure, to resolve the static indeterminateness of reactions on an inclined beam is likely to be inaccurate.

Measurements were also made of the impact loading on a ladder by the ascent and descent of a climber. In a typical ascent without hesitation, the impact loading

was found to be about 1 1 2 times the climber’s weight. This result is consistent

with the predicted maximum value of 2 for the ratio of impact load to static load by a dynamic model of the ladder as a linear elastic medium subject to a pulse perturbation.

The successful description of the reaction forces on a ladder modeled as a single beam prompted the question of whether any of the three theoretical models satisfactorily accounted for the reaction forces on an actual single beam. Measurements analogous to those made on the ladder were also made on a wood beam of comparable length. The results showed that the same model that accounted for the reaction forces on the ladder also best predicted the reaction forces on the single wood beam. Whether this result is completely general, or whether a different outcome might arise for a beam of different composition, length, and/or cross section can only be determined by further experimentation.

A final point of importance concerns the inclusiveness of the three fundamental models. In other words, is there another independent way to analyze an Euler-Bernoulli (E-B) beam ladder that might produce better agreement with experiment than Model 3 of this paper? In the author’s opinion, the answer is probably “no”. The reason is that other models within the single E-B beam framework are either irrelevant, equivalent, or inconsistent.

For example, two models not analyzed in this paper include a ladder with simple (i.e. frictionless) support at the wall and a ladder with fixed support (i.e. infinite friction) at the wall. Neither model is relevant because the contact at the wall is neither frictionless nor immovable. (The models of this paper all assume non-slippage at the ground, for otherwise there would be no static equilibrium.)

Consider, however, a variation of Model 1. Model 1 resolved static indeterminateness by applying two boundary conditions to the wall support in the configuration of Panel B of

Another conceivable model might be to express U in terms of R 3 and R 4 but impose only one boundary condition at the wall, e.g. no displacement vertically. The resulting equation, which couples the vertical and horizontal reactions at the wall, is then used together with Equation (31) which expresses zero torque in static equilibrium. This procedure is inconsistent, however, because the first equation leads to a rotational moment at the ground support, whereas the second equation describes the absence of such a moment. The resulting set of reactions depends on the length (L) and width (h) of the beam (in conflict with experimental results) and leads to large reaction forces that violate physical laws.

In conclusion, for any model of a ladder (or other physical structure) it is paramount that the equations of static equilibrium and the complementary relations be consistent and compatible with the boundary conditions.

E. Donovan, W. Emery, J. Hallquist, W. Scully, L. Sewanan, and W. Strange each helped take data at some point during the initial phase of this project. The author also thanks Trinity College for partial support through the research fund associated with the George A. Jarvis Chair of Physics.

The author declares no conflicts of interest regarding the publication of this paper.

Silverman, M.P. (2018) Reaction Forces on a Fixed Ladder in Static Equilibrium: Analysis and Definitive Experimental Test of the Ladder Problem. World Journal of Mechanics, 8, 311-342. https://doi.org/10.4236/wjm.2018.89024

The Heaviside function defined in Equation (12) and repeated below

H ( x − x 0 ) = { 0 x < x 0 1 x ≥ x 0 (54)

provides an economic way to represent piecewise continuous functions, such as frequently occur in the mechanics of continuous media. In the representation of static forces, impact forces, rotational moments, and strain energy densities it is often necessary to integrate over expressions of the form z n H ( z − a ) where a is a non-negative constant and n is a non-negative integer. The general result is

∫ 0 x z n H ( z − a ) d z = ( x n + 1 − a n + 1 n + 1 ) H ( x − a ) ( n = 0 , 1 , 2 , ⋯ ) . (55)

A general proof, which will not be given here, can be constructed inductively from the examples below for n = 0 , 1 .

1) Case n = 0 :

∫ 0 x H ( z − a ) d z = ∫ 0 x H ( z − a ) d ( z − a ) = ∫ − a x − a H ( u ) d u = ( x − a ) H ( x − a ) (56)

Steps: Transform the integration variable from z to u = z − a and use the properties of H ( u ) in Equation (54).

2) Case n = 1 :

∫ 0 x z H ( z − a ) d z = ∫ 0 x ( z − a ) H ( z − a ) d ( z − a ) + a ∫ 0 x H ( z − a ) d ( z − a ) = ∫ − a x − a u H ( u ) d u + a ∫ − a x − a H ( u ) d u = ( x − a ) 2 2 H ( x − a ) + a [ ( x − a ) H ( x − a ) ] = x 2 − a 2 2 H ( x − a ) (57)

Steps:

・ Line 1: Transform the integrand into a function of u = z − a by appropriate subtraction and addition of terms.

・ Line 2: Perform the first integral and import the result obtained in the case n = 0 for the second integral.

・ Lines 3 and 4: Expand and simplify the algebraic expressions to obtain the form of Equation (55).