Department of Civil Engineering, ISM DhanbadStructural Analysis-I Practical Manual

Deflection of Curved MembersAim of the experiment: To determine the elastic displacement of the curved members experimentally and verification of the same by theoretical results.Apparatus required: Various curved members, viz. quadrant of a circle, semicircle with straight arm, quadrant of a circle with straight arm and circle, dial gauge, weights, scale and etc. Theory: Castigliano's first theorem is used to find the elastic displacements of curved members. Theorem states that the partial derivative of the total strain energy of a linearly elastic structure expressed in terms of displacements with respect to any displacement j at coordinate j is equal to the force Pj at coordinate j. The theorem may be expressed symbolically as

In all cases the horizontal B,horizontal and vertical deflection B,vertical due to vertical load W are to be determined. These deflections are obtained by using Castigliano's first theorem where strain energy due to bending only is taken into account. The results obtained for the four curved members is given as below - a) Quadrant of a circleFixed at A and free at B (having radius R) and subjected to a concentrated load W at free end.

RAWB

Figure 1.a

Vertical displacement of load = B, vertical =

Horizontal displacement of load = B, horizontal = b) Semicircle with straight armFrom A to B is a semi circle of radius R, B to C is a straight length of y

yRWCBA

Figure 1.bVertical displacement of loaded point C

C, vertical =

Horizontal displacements of loaded point C = C, horizontal = c) Quadrant with a straight legFrom A to B is a quadrant of a circle of radius R and from B to C, straight length of y.

A

yWRWWCB

Figure 1.c

Vertical displacement of load point A = A, vertical =

Horizontal displacement of load point A = A, horizontal = d) WCircle of radius R

Figure 1.dRBA

Vertical displacement of loaded point B = B, vertical Procedure:1. Place a load on the hanger to activate the member and treat this as the initial position for measuring deflections.2. Fix the dial gauges for measuring horizontal and vertical deflections.3. Place the additional loads at the steps mentioned in the table below for each case and tabulate the values of dial gauge reading against the applied loads.4. Plot the graph load Vs deflection for each case to show that the structure remains within the elastic limit.5. Measure the value of R and straight length in each case. Find width and depth of steel section and calculate the value of I as bd3/12.

Observations and table:Width of section (mm) b=Depth of section (mm) d=

Least moment of inertia= E (N/mm2)= 2 x 106.

(a) Quadrant of a circleSl. No.Additional load (kg)Dial gauge reading (mm)Deflection (mm)

Horizontal directionVertical directionHorizontal directionVertical direction

(b) Quadrant with Straight legSl. No.Additional load (kg)Dial gauge reading (mm)Deflection (mm)

Horizontal directionVertical directionHorizontal directionVertical direction

(c) Semi-circle with straight legSl. No.Additional load (kg)Dial gauge reading (mm)Deflection (mm)

Horizontal directionVertical directionHorizontal directionVertical direction

(d) CircleSl. No.Additional load (kg)Dial gauge reading (mm)Deflection (mm)

Horizontal directionVertical directionHorizontal directionVertical direction

Results:

Prepared by Puja rajhans