university solved problems on overhanging beam

UNIVERSITY SOLVED PROBLEMS ON OVERHANGING BEAM

Example 2.42:

Draw the shear force and bending moment diagrams for the beam shown in Fig. 2.55(a) indicating principle values.

Given: 

As shown in Fig.2.55(a).

To draw: 

SFD and BMD

Solution: 

Taking moment about A,

SF calculation:

SF at D = 0

SF at C (without reaction RC)

SF at C (with reaction RC)

= 6.25 - RC = 6.25 - 61.9 = - 55.65 kN

SF at B (without point load)

= −55.65 + 10 × 2 = - 35.65 kN

SF at B (with point load)

= -35.65 + 20 = 15.65 kN

SF at A = RA = 44.35 kN

Join the values CD by parabolic curve and all other values by inclined straight lines as shown in Fig.2.55(b).

BM calculation:

BM at D = 0

BM at A = 0

Join all the values AB and C by parabolic curves and CD by a cubic curves as shown in Fig.2.55(c).

Result: 

The SFD and BMD are shown in Fig.2.55(b) & (c) respectively.

Example 2.43: 

Draw the shear force and bending moment diagrams for the beam loaded as shown, locate the point of contraflexure and find the maximum and minimum values of BM.

Given: 

As shown in Fig.2.56(a).

To find:

(i) Draw SFD and BMD.

(ii) Find the maximum and minimum of BM.

Solution: 

Taking moment about B,

RC = 33.375 kN

RB + RC = 6 × 13 = 78 kN

RB = 78 - 33.375 = 44.625 kN

SF calculation:

SF at D = 0

SF at C

(without reaction RC)

= 6 × 2 = 12 kN

SF at C (with reaction RC)

= 6 × 2 -33.375 = -21.375 kN

SF at B (without reaction RB)

= 6 × 10 - 33.375 = 26.625 kN

SF at B (with reaction RB)

= 6 × 10 - 33.375 - 44.625

= -18 kN

SF at A = 6 × 13 - 33.375 - 44.625 = 0 

Join all the values as shown in Fig.2.56(b).

BM calculation:

BM at D = 0

BM at A = 0

Maximum BM calculation:

Taking the SF at a distance 'x' m from D.

SFx = 6 × x - 33.375 = 0

x = 5.5625 m from D 

Maximum BM:

Join all the values by parabolic curves as shown in Fig.2.56(c).

Result: 

The SFD and BMD are shown in Fig.2.56(b) & (c) respectively.

Minimum value of BM = 0

Maximum value of BM = 26.07 kN-m

Example 2.44:

Draw the SF and BM diagrams for the beam shown below, Find the maximum values and their positions. Give the values at important points in the diagram.

Given: 

As shown in Fig.2.57(a).

To draw:

(i) SFD and BMD

(ii) Maximum BM

Solution: 

Taking moment about A,

RA + RB = 2 × 6 + 2 = 14 kN

RA = 14 - RB = 14 - RB = 14 - 12 = 2 kN

SF calculation

SF at C = 2 kN

SF at B (without Reaction RB) = 2 +2 × 2 = 6 kN

SF at B (with Reaction RB) = 6 - RB

= 6 – 12 = -6 kN

SF at A = RA = 2 kN

Join all the values as shown in Fig.2.57(b).

BM calculation:

BM at C = 0

BM at A = 0

Maximum BM calculation:

Taking SF at a distance 'x' m from C

SFx = 2 – 12 + 2 × x = 0

-10 + 2 x = 0

⇒ 2x = 10

x = 5 m from C

Maximum BM:

Join all BM values as shown in Fig.2.57(c).

Result: 

The SFD and BMD are shown in Fig.2.57(b) & (c) respectively. 

The maximum BM 1 KN-m occurs at a distance of 5 m from end C.

Example 2.45: 

Draw the shear force and bending moment diagram for the loaded beam shown in Fig.2.58(a).

Given: 

As shown in Fig.2.58(a).

To draw: 

SFD and BMD

Solution: 

Taking moment about A,

= 56.875 kN

SF calculation:

SF at D = 0

SF at B (without reaction RB)

= 10 × 3 = 30 kN

SF at B (with reaction RB)

= 30 - 73.125 = - 43.125 kN

SF at A (without reaction RA)

= 10 × 11 - 73.125 = 36.875 kN

SF at A (with reaction RA)

= 36.875 - 56.875 = - 20 kN

SF at C = 0

Join all the values as shown in Fig.2.58(b).

BM calculation:

BM at D = 0

BM at C = 0

Maximum BM calculation:

SF equation where it changes sign. (i.e., x m from the end D),

SFx = 10 × x - RB

10 × x - 73.125 = 0

x = 7.3125 m

Therefore,

= 47.988 KN-m

Join all the BM values as shown in Fig.2.58(c).

Result: 

The SFD and BMD are shown in Fig.2.58(b) & (c).

Example 2.46: 

A beam 6m long rests on supports 5m apart, the right hand end is overhanging by Im. The beam carries a UDL of 20 kN/m over the entire length of the beam. Draw SFD and BMD indicating the maximum BM and the point of contraflexure. 

Given: 

As shown in Fig.2.59(a).

To draw:

(i) SFD and BMD

(ii) Mmax and Point of contraflexure

Solution: 

Taking moment about A,

RA + RB = 20 × 6 = 120 kN

RA = 120 - RB = 120 - 72 = 48 kN

SF calculation:

SF at C = 0

SF at B (without reaction RB)

= 20 × 1 = 20 kN

SF at B (with reaction RB)

= 20 – 72 = -52 kN

SF at A = RA = 48 kN

Join all the values as shown in Fig.2.59(b).

BM calculation:

BM at C = 0

BM at A = 0

Maximum BM:

Consider the distance x m from C where the SF changes its sign.

Join all BM values as shown in Fig.2.59(c).

Point of contraflexure:

Point of contraflexure is the point where BM is zero. Consider the point which is distance from C Where BM is zero.

Taking moment about that point,

y = 6 is not a possible value.

⸫ The point of contraflexure is 1.2m from C.

Result:

(i) The SFD and BMD are shown in Fig.2.59(b) & (c) respectively.

(ii) Maximum value of BM = 57.6 kN-m

(iii) The point of contraflexure occurs at a distance of 1.2 m from end C. 

Example 2.47: 

Draw the SFD and BMD for the beam loaded as shown in Fig. 2.60.

Given: 

As shown in Fig.2.60(a).

To draw: 

SFD and BMD

Solution: 

Taking moment about A,

SF calculation:

SF at C = 0

SF at B (without reaction RB)

= 12 × 2 = 24 kN

SF at B (with reaction RB)

= 12 × 2 - RB

= 24 - 64 = -40 kN

SF at A= +RA = +32 kN

Join all the SF values by inclined line as shown in Fig.2.60(b).

BM calculation:

BM at C = 0

BM at A = 0

Join the above BM values by parabolic curve as shown in Fig.2.60(c).

Result: 

The SFD and BMD are shown in Fig.2.60(b) & (c).

Example 2.48:

Construct the SFD and BMD for the beam shown in Fig. 2.61(a).

Given: 

As shown in Fig.2.61(a).

To draw: 

SFD and BMD

Solution: 

Taking moment about A,

= 6 + 30 + 25 = 61

= 6 + 10 + 3.75 = 19.75 KN

RA = 19.75 - RD = 19.75 - 12.2

RA = 7.55 kN

SF calculation:

SF at E = 0

SF at D (without reaction)

SF at D (with reaction RD)

= 3.75 - RD = 3.75 - 12.2

= - 8.45 kN

SF at C = -8.45 + 10 = 1.55 kN

SF at B = 1.55 kN

SF at A = RA = 7.55 kN

Connect the values between D and E by parabolic curve and all other values are joined as shown in Fig.2.61(b).

BM calculation:

BM at E = 0

BM at A = 0

Join the values between C and E by cubic curve and AB by parabolic curve as shown in Fig.2.61(c).

Result: 

The SFD and BMD are drawn as shown in Fig.2.61(b) & (c) respectively.

Example 2.49:

Draw the SFD and BMD for the beam shown in Fig. 2.62(a).

Given: 

As shown in Fig.2.62(a).

To draw: 

SFD and BMD

Solution: 

Taking moment about B,

RB + RD = 1.5 × 2 + 2 × 2 + 6 = 13 kN

RB = 13 - RD = 13 - 8.75 = 4.25 kN

SF calculation:

SF at E = 0

SF at D (without reaction RD) 2 × 2 = 4 kN

SF at D (with reaction RD)

= 4 - RD =  4 - 8.75 = -4.75 kN

SF at C = -4.75 + 6 = 1.25 kN

SF at B (without reaction RB) = -4.75 + 6 = 1.25 kN

SF at B (with reaction RB) = 1.25 - RB

= 1.25 - 4.25 = -3 kN

SF at A = 0

Join all SF values as shown in Fig.2.62(b).

BM calculation:

BM at E = 0

BM at A = 0

Join all the BM values as shown in Fig.2.62(c).

Result: 

The SFD and BMD are as shown in Fig.2.62(b) & (c) respectively.