Consider the free body diagram of all forces at B.

Q1  = tan -1   = 51.34º

Q2 = tan-1 = 45º

Let the forces developed in the member AB & AC are FAB &FBCrespectively. The force diagram will be as follows at B.

Applying Lami’s Theorem:

FAB = 1067.19 N

FBC = 942.81 N

The string passes over the pulley & is attached to weight w = 400 N.  

Thus TCD = TDE = 400 N

Consider joint C & apply Lami’s theorem

Now consider FBD at joint B

By using Lami’s theorem

TAB = 230.94 N

) Consider the free body diagram of sphere B as shown below:

By lami’s theorem

RRR = 939.69 N

  = RAB

Now consider FBD of sphere A

∑Fx = 0

-RB cos 20 – RQ cos 70 + Rp = 0

-342.02 cos 20 – RQ sin 70 + Rp  = 0

Rp = 321.39 + 0.34 RQ          ------------------(1)

∑ fy = 0

RQ sin 70 – 342.02 sin 20 – 1000 = 0

RQ = 1188.66 N    - put in equation (1)

Rp = 725.5 N

Consider the free body diagram of joint B as shown in figure

Assume that member BC is in compression & AB is in tension. By using Lami’s theorem at B,

FAB = -740.33 kN 

-ve sign indicates that member AB is not in Tension but is in compression.

Now,

-ve sign indicates that member BC is not in compression but in tension.

Consider the FBD of sphere B as shown in figure above.

Applying lami’s theorem, we have

RC = 190.84 N

TAB = 101.54 N

Consider the FBD as shown

Applying conditions of equilibrium,

∑fx = 0

RHA – TCD = 0

RHA = TCD ------- (1)

∑fy = 0

RVA – 500 = 0

RVA = 500 kN

Taking moment about point A,

∑MA = 0

-(TCD × 5 ) + (500 × 9) = 0

-TCD × 5 = -4500

TCD =

TCD = 900 kN

Putting the value in equation (1)

RHA = TCD

RHA = 900 kN

Consider triangle AEC

EC = 1.61 m

Consider triangle AEC

AE = 1.92 m

Applying conditions of equilibrium

∑fX = 0

-TBD cos15 + RHA = 0 -------- (1)

∑fy = 0

RVA – 200 – TBD sin15 = 0

RVA – 200 – 0.26TBD = 0 ---------(2)

Taking moment at A

∑MA = 0

-(TBD cos15 × 3.2) + (TBD sin15 × 3.84) + (200 × 1.92) = 0

-3.09 TBD + 0.99 TBD + 384 = 0

-2.1 TBD = -384

TBD = 182.86 N              ---put in (1) & (2)

RHA = TBD cos15 = 182.86 cos15

RHA = 196.62 N

RVA = 200 + 0.26TBD = 200 + (0.26 × 182.86)

RVA = 247.54 N

1. Roller Support: -

Roller support is show in two ways as shown:

• A free body diagram of roller support is shown in figure given below

• Roller support offers only one reaction; which is always perpendicular to the base or plane of roller.

• All the steel trusses of the bridges have one of their ends supported on roller.

• Main advantages of the support are that beam can easily move towards left or right due to expansion & contraction.

Types of Beam: -

1. Simply supported beam –

A beam which is just resting on the supports at the end without any connection is known as simply supported beam. It is generally used for vertical landing system.

2.     Overhanging Beam –

A beam which is supported at the intermediate point other than ends is called as overhanging beam. Here portion of beam is extended beyond the support

a)

       (singly overhanging beam)

b)

(doubly overhanging beam)

3.     Cantilever Beam –

A beam which is fixed at one end is called as cantilever beam

Here, there are three reactions components:

1. Vertical reaction at A (RVA)
2. Horizontal reaction at A (RHA)
3. Fixing moment at A (MA)

We can assume any direction for above components.

4.     Continuous Beam:

A beam having more than two support is called as continuous beam.

5.     Compound Beam:

When two or more beams are joined together by using internal hinge; or when one beam rests over another beam by using internal roller, then such beam is called as compound beam.