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2.11. PROBLEMS FOR EQUILIBRIUM OF RIGID BODIES INTWO•DIMENSIONAND THREE DIMENSIONS

2.11.1. Problems on simple supported beam

Problem 2.18. A simply supported beam AB of span 6 m carries point loads of 3 kN and
6 kN at a distance of
2 m and 4 m from the left end A as shown in Fig. 2.39. Find the reactions
at A and B analytically and graphically.

Sol. Given:

Span of beam = 6 m

Let RA = Reaction at A

RB = Reaction atB 

Fig.2.39

(a)    Analytical method. As the beam is in equilibrium, the moments of all the forces about any point should be zero.

Now taking the moment of all forces about A, and equating the resultant moment to zero, we get

(a)   Graphical method. First of all draw the space diagram of the beam to a suitable

scale. Let 1 cm length in space diagram represents 1 m length of beam. Hence take AB = 6 cm, distance of load 3 kN from A = 2 cm and distance of 6 kN from A = 4 cm as shown in Fig. 2.40 (a).

Now name all the loads and reactions according to Bow’s notation i.e., load 3 kN is named by PQ, load 6 kN by QR, reaction RB by SR and reaction RA by SP.

Now the vector diagram is drawn according to the following steps: [Refer to Fig. 2.40 (b)].

  1. Choose a suitable scale to represent various loads. Let 1 cm represents 1 kN load. Hence load PQ (i.e., 3 kN) will be equal to 3 cm and load QR (i.e., 6 kN) 6 cm.
  2. Now take any point p and draw line pq parallel to load PQ (i.e., 3 kN). Take pq = 3 cm to represent the load of 3 kN.

Through q, draw line qr parallel to load QR (i.e., 6 kN). Cut qr equal to 6 cm to represent the load of 6 Fig.2.40(a)

Fig.2.40 (a) Space diagram

Fig.2.40(b)

Fig.2.40 (b) Vector diagram

4.   Now take any point O. Join the point O to the points p, q and r as shown in Fig. 2.40 (b).

5.   Now in Fig. 2.40 (a), extend the lines of action of the loads (3 kN and 6 kN), and the two reactions. Take any point 1, on the line of action of the reaction RA. Through 1, draw the line 1-2 parallel to pO, intersecting the line of action of load 3 kN at point 2.

6.  From point 2, draw line 2-3 parallel to qO, intersecting the line of action of load 6 kN at 3. Similarly, from point 3, draw a line 3-4 parallel to rO, intersecting the line of action of reaction RB at point 4,

7.  Join 1 to 4. The line 1-4 is known as closing line. From the vector diagram, from point O, draw line Os parallel to line 1-4.

8. Measure the length sp and rs. The length sp represents the reaction RA and length rs represents the reaction RB.

By measurement, sp = 4 cm and rs = 5 cm

RA = Length sp × scale = 4 × 1 kN = 4 kN.   Ans.

RB = Length rs × scale = 5 × 1 kN = 5 kN.   Ans.

Problem 2.19A simply supported beam AB of length 9 m, carries a uniformly distributed load of 10 kN/m for a distance of 6 m from the left end. Calculate the reactions at A and B.

Sol. 

Given :

Length of beam = 9 m

Rate of U.D.L. = 10kN/m

Length of U.D.L. = 6 m

Total load due to U.D.L.

= (Length of U.D.L.) × Rate of U.D.L.
= 6 × 10 = 60 kN 

The load of 60 kN will be acting at the middle point of AC i.e., at a distance of    3 m from A.

Let.            RA = Reaction at A       and       RB = Reaction at B

Taking the moments of all forces about point A, and equating the resultant moment to zero, we get

Problem 2.20. A simply supported beam of length 10 m, carries the uniformly distributed
Load and two point loads as shown in Fig. 2.42. Calculate the reactions RA and RB·

Sol. Given:

Length of beam
Length of U.D.L.
Rate of U.D.L,

Total load due to U.D.L. = 4 × 10 = 40 kN

= 2 m from C ( or at a distance of  2 + 2 = 4 m from point A).

Let       RA = Reaction at A

and      RB = Reaction at B

Taking the moments of all forces about point A and equating the resultant moment to zero, we get

Problem 2.21. A simply supported beam of span 9 m carries a uniformly varying load from zero at end A to 900 N/m at end B. Calculate the reactions at the two ends of the support.

Sol. Given:

Span of beam = 9 m

Load at end A = 0

Load at end B = 900 N/m

Also for equilibrium of the beam, Fy = 0

or                RA + RB = Total load on beam = 4050

          RA = 4050 – RB = 4050 – 2700 = 1350 N. Ans.

Problem 2.22. A simply supported beam of length 5 m carries a uniformly increasing load of 800 N/ m at one end to 1600 N/m at the other end. Calculate the reactions at both ends.

Sol. Given:

Length of beam = 5 m

Load at A = 800 N/m

Load at B = 1600 N/m

Total load on the beam = Area of load diagram ABDC

= Area of rectangle ABEC + Area of ΔCED

= 4000 + 2000 = 6000 N

Let RA = Reaction at A

and RB = Reaction at B.

Taking the moments of all forces about point A and equating the resultant moment to zero, we get

RB × 5 – (Load due to rectangle) × Distance of C.G. of rectangle from A
- (Load due to triangle) × Distance of C.G. of triangle from A = 0

2.11.2. Problems on Overhanging Beams. If the end portion of a beam is extended beyond the support, then the beam is known as overhanging beam. Overhanging portion may be at one end of the beam or at both ends of the beam as shown in Fig. 2.45.

Problem 2.23. A beam AB of span 8 m, overhanging on both sides, is loaded as shown in Fig. 2.46. Calculate the reactions at both ends.

Sol. Given:

Span of beam = 8 m
Let RA = Reaction at A
and RB = Reaction at B.

Taking the moments of all forces about point A and equating the resultant moment to zero, we get

RB × 8 + 800 × 3 – 2000 × 5 – 1000 × (8 + 2) = 0

or       8RB + 2400 – 10000 – 10000 = 0

or       8RB = 20000 – 2400 = 17600

Also for the equilibrium of the beam, we have

RA + RB = 800 + 2000 + 1000 = 3800

RA = 3800 – RB = 3800 – 2200 = 1600 N. Ans.

Problem 2.24. A beam AB of span 4 m, overhanging on one side upto a length of 2 m, carries a uniformly distributed load of 2 kN/m over the entire length of 6 m and a point load of 2 kN/m as shown in Fig.2.47. Calculate the reactions at A and B. 

Sol. Given:

Span of beam = 4 m

Total length = 6 m

Rate of U.D.L.            = 2 kN/m

Total load due to U.D.L. = 2 × 6 = 12 kN
The load of 12 kN (i.e., due to U.D.L.) will act at the middle point of AC, i.e., at a distance of 3 m from A,

Let RA = Reaction at A

and        RB = Reaction at B.

Taking the moments of all forces about point A and equating the resultant moment to zero, we get

RB × 4 – (2 × 6) × 3 – 2 × (4 + 2) = 0
or     4RB – 36 -12 = 0

or     4RB = 36 + 12 = 48

2.11.3. Problems on Roller and Hinged Supported Beams. In case of roller supported beams, the reaction on the roller end is always normal to the support. All the steel trusses of the bridges is generally having one of their ends supported on rollers. The main advantage of such a support is that beam, due to change in temperature, can move easily towards left or right, on account of expansion or contraction.

In case of a hinged supported beam, the reaction oil the hinged end may be either vertical inclined, depending upon the type of loading. The main advantage of a hinged end is that the beam remains stable. Hence all the steel trusses of the bridges, have one of their end on rollers and the other end as hinged.

Problem 2.25. A beam AB 1.7 m long is loaded as shown in Fig. 2.48. Determine the reactions at A and B.

Sol. Given:

Length of beam = 1. 7· m

Let       RA = Reaction .at A
and   RB = Reaction at B.

Since the beam is supported on rollers at B, therefore the reaction RB will be vertical. The beam is hinged at A, and is carrying inclined load, therefore the reaction RA will be inclined. This means reaction RA will have two components, i.e., vertical component and horizontal component.

Let       RAX = Horizontal component of reaction RA
RAY = Vertical component of reaction R A’

First resolve all the inclined loads into their vertical and horizontal components.
(i) Vertical component of load at D

= 20 sin 60° = 20 × 0.866 = 17.32 N

and its horizontal component

= 20 cos 60° = 10 N ←
(ii) Vertical component of load at E

= 30 sin 45° = 21.21 N

and its horizontal component

= 30 cos 45° = 21.21 N →
(iii) Vertical component of load at B

= 15 sin 80° = 14.77 N

and its horizontal component

= 15 cos 80° = 2.6 N ←
From condition of equilibrium, Fx = 0
or        RAX - 10 + 21.21- 2.6 = 0

or         RAX = 10 – 21.21 + 2.6 = – 8.61 N

-ve sign shows that the assumed direction of RAX (i.e., horizontal component of RA) is wrong. Correct direction will be opposite to the assumed direction. Assumed direction of RAX is towards right. Hence correct direction of RAX will be towards left at A.

RAX = 8.61 N ←

To find RB, take moments* of all forces about A.

For equilibrium, MA = 0

Problem 2.26. A beam AB 6 m long is loaded as shown in Fig. 2.49. Determine the reactions atA and B by (a) analytical method, and (b) graphical method.

Sol. Given:

Length of beam = 6 m

Let RA = Reaction at A
RB = Reaction at B.

The reaction RB will be vertical as the beam is supported on rollers at end B. 

The reaction RA will be inclined, as the beam is hinged at A and carries inclined load.

Let RAX = Horizontal component of reaction RA
RAY = Vertical component of reaction RA.

(a) Analytical method. First resolve the inclined load of 4 kN into horizontal and vertical components.

Horizontal component of 4 kN at D

= 4 cos 45° = 2.828 kN →

and its vertical component

= 4 sin 45° = 2.828 kN ↓

For equilibrium,

*The moment of horizontal component about A, will be zero.

(b) Graphical method. First of all convert the uniformly distributed load (U.D.L.) into its equivalent point load acting at the C.G. of the portion on which U.D.L. is acting. Hence total load due to U.D.L. will be 1.5 × 2 = 3 kN acting at a distance of 3 m from point A

 i. Now draw the space diagram of the beam according to some suitable scale, as shown in Fig. 2.50(a).

 ii. Name all the loads and reactions according to Bow’s notation. Now draw the vector diagram as shown in Fig. 2.50(b). Choose any suitable scale for vector diagram.

 iii.Take any point p for drawing vector diagram. From p, draw line pq parallel and equal to load 5 kN (i.e., load PQ). Fromq, draw qr parallel and equal to 3 kN. From r, draw rs parallel and equal to 4 kN load.

iv.Now take any point O, and join Op, Oq, Or and Os.

v. Now in space diagram [i.e., Fig. 2.50(a)], extend the lines of actions of loads PQ, QR, RS and reaction RB.

vi.Take any point 1, vertically below the point A as shown in Fig. 2.50(a). From point 1, draw line 1-2 parallel to line pO, intersecting the line of action of 5 kN at point 2.

 vii. Similarly, draw lines 2-3, 3-4 and 4-5 parallel to qO, rO and sO respectively. Join point 1 to 5. Line 1-5 is the closing line in space diagram.

 viii. From O in vector diagram, draw a line parallel to closing line 1-5. Now through 8, draw a line st vertical (as the reaction RB is vertical), intersecting the line through O at t. Join t to p.

 ix.  The length st represents the reaction RB in magnitude and direction whereas the length tp gives the magnitude and direction of reaction RA. At point A, draw a line parallel to tp as shown in Fig.2.50(a). By measurement, we get

RA = length tp = 6.43 kN
RB = length st = 5.052 kN
and     θ = 26.1o

Fig.2.50 (a) Space diagram

Fig.2.50 (b) Vector diagram

Problem 2.27. A beam AB 10 m long is hinged at A and supported on rollers over a smooth surface inclined at 30° to the horizontal at B. The beam is loaded as shown in in Fig. 2.51. Determine reactions at A and B.

Sol. Given :

Length of beam = 10m

Let RA = Reaction at A

and RB = Reaction at

The reaction RB will be normal to the support as the beam at B is supported on the rollers. But the support at B is making an angle 30o with the horizontal or 60o with the vertical as shown in Fig. 2.51. Hench the reaction of RB is making an angle of 30o with the vertical.

The component of RB

= RB cos 30°
and horizontal component of RB

= RB sin 30°

These components are shown in Fig. 2.51(a). 

Resolving the load of 5 kN acting at D into horizontal and vertical components, we get

Vertical component of 5 kN

= 5 sin 45° = 5 × 0.707 = 3.535 kN

Horizontal component of 5 kN

= 5 cos 45° = 5 × 0.707 = 3.535 kN

The reaction at A will be inclined, as the end A is hinged and beam carries inclined load.

Let      RAX = Horizontal component of reaction RA

RAY = Vertical component of reaction RA

For equilibrium of the beam, the moments of all forces about any point should be zero.

Taking the moments* about point A,

(RB cos 30°) × 10 – 4 × 2.5 – (5 sin 45°) × 5- 5 × 8 = 0.

8.66 RB -10 -17.675 – 40 = 0

For equilibrium, ƩFx = 0

or    RAH + 5 cos 45° - RB sin 30° = 0

or    RAH + 3.535 -7.81 × 0.5 = 0

      RAH = 7.81 × 0.5 – 3.535 = 0.37 kN

For equilibrium, ƩFy = 0

    RAV + RB cos 30° – 4 – 5 sin 45° – 5 = 0

or    RAV + 7.81 × 0.866 – 4 – 3.535 – 5 = 0

or    RAV + 6.763 – 12.535 = 0

or    RAV= 12.535 – 6.763 = 5.77 kN

 

 

                = 5.78 kN Ans.

The angle made by RA with x-direction is given by

 

 

θ = tan-1 15.59 = 86.33°. Ans.

* The moments of horizontal components of 5 kN at D of reaction RB will be zero about the point A.

Problem 2.28. Determine the reactions at the hinged support A and and the roller support B as shown in Fig. 2.52(a).

Sol. Given:

The support at A is hinged whereas the support at B is placed on the roller. Hence the reaction at the roller support will be perpendicular to the inclined surface.

The reaction at the hinged support A (i.e., reaction RA) will be inclined at some angle to
the inclined surface AB.

Let RAV = Component of reaction RA normal to inclined surface AB

RAH = Component of reaction RA along the inclined surface AB.

The given vertical force of 500 N at C is resolved parallel and perpendicular to the inclined surface AB. The value of its component parallel to the inclined surface is equal to 500 sin 30° and its value perpendicular to the inclined surface is equal to 500 cos 30° or 250 × √3 N. The directions of these components are shown in Fig. 2.52(b).

Similarly, the vertical force of 500 N at D is resolved as shown in Fig. 2.52(b).

Now sum of the components parallel to the inclined surface AB = 500 sin° + 500 sin 30° = 250 + 250 = 500 N acting from left to right.

Hence the value of the reaction component RAH would be equal to 500 N and should act from right to left as shown.

To find RB, take the moments of all forces about point A shown in Fig. 2.52 (a).

Hence applying ƩMA = 0,we get

Fig. 2.52

500 × AC + 500 × AD = RB × AB

But AC = 5 cm, AD = 10 cm

 

 

 

 

Hence above equation becomes,

500 × 5 + 500 × 10 =RB × 13 or 2500 + 5000 =RB × 13

 

 

 

Now from Fig. 2.52 (b), equating all the forces perpendicular to the inclined surface AB, we get

RAV + RB = 500 × cos 30° + 500 cos 30°

 

 

 

 

 

 

 

 

Problem 2.29. Find reactions at supports of an L-bent shown in Fig. 2.53.

Sol. Given:

Force at point D = 100 N at an angle of 30owith horizontal

Force at point C = 70 N at an angle of 45o with vertical

Load on EF = 250 N/m = 250 × Length EF in meter

= 250 × 0.6 = 150 N

The load on EF will be acting at the middle point of EF i.e., at a distance of 0.6/2 =0.3 m from E or at a distance of 0.6 + 0.3 = 0.9 m from A, The point B is placed on roller at an angle of 20o with the horizontal. Hence reaction at B will be normal to the surface of the roller.

The perpendicular distance from A on the line of action of RB =AO =AB cos 20°. 180 × cos cos 20° cm = 1.8 cos20° cm as shown in Fig. 2.53(a). For equilibrium of the beam, the moments of all forces about any point should be zero. Taking moments of all forces about point A, we get

[Horizontal component at D] × AD - [Horizontal component at C] × AC + Load on EF × 90 – RB × AO = 0

or (100 cos 30) × 80 – (70 × sin 45) × 40 + 150 × 90 – RB × 180 cos 20 = 0

(Note: The vertical components at D and C, pass through the point A. Hence moments of these vertical components about A are zero).

or 6928 – 1979.6 + 13500 – 169.14 RB = 0

or 18448.4 = 169.14 RB

 

 

Let RA = Reaction at the point A

The reaction at A can be resolved in two components i.e., RAX and RAY

For equilibrium, ƩFx = 0

or RAX + 100 cos 30° – 70 sin 45° - RB sin 20 = 0

RAX = RB sin 20° + 70 sin 45° – 100 cos 30°

= 109.07 × 0.342 + 70 × 0.707 -100 × 0.866

= 37.3 + 49.49 – 86.6 = 0.19 N

For equilibrium, ƩFy = 0

or RAX +100 sin 30° + 70 cos 45° + RB cos 20° = 150

or RAY = 150 – 100 sin 30° + 70 cos 45° + RB cos 20°

= 150 – 50 – 49049 – 109.07 × 0.9396 = -51.98 N

(-ve sign means, RAy will be acting vertically downward)

Refer to Fig. 2.53(c) 

 

 

 

 

 

 

 

 

 

 

 

2.11.4. Problems when Beams are Subjected to Couples. In this section, the reactions of the beam will be calculated when beams are subjected to clockwise or anti-clockwise couple along with the other loads. While taking the moments about any point, the magnitude and sense of the couple is taken into consideration. But when the total load on the beam is calculated the magnitude and sense of the couple is not considered.

Problem 2.30. A simply supported beam AB of 7 m span is subjected to : (i) 4 kN m clockwise couple at 2 m from A, (ii) 8 kN m anti-clockwise couple at 5 m from A and (iii) a triangular load with zero intensity at 2 m from A increasing to 4 kN per m at a point 5 m from A. Determine reactions at A and B.

Sol. Given:

Span of beam = 7 m

Couple at C (i.e., at 2 m from A) = 4 kN m (clockwise)

Couple at D (i.e., at 5 m from A) = 8 kN m (anti-cloakwise)

Triangular load from C to D with:

Vertical load at C = 0

Vertical load at D = 4 kN/m

 

or 2+2+4 m from end A.

Let RA = Reaction at A.

RB = Reaction at B.

Taking the moments of all forces about point A and equating the resultant moment to
zero (i.e.,) ƩMA = 0 and considering clockwise moment positive and anti-clockwise moment negative), we get

-RB × 7 + 4* – 8** + (Total load on beam) × (Distance of total load from A) = 0
or  -7 RB + 4 – 8 + 6 × 4 = 0

- 7 RB + 4 – 8 + 24 = 0

or 20 = 7RB