Consider the frame and loading shown in the adjoining figure. The temerature act
ID: 1711329 • Letter: C
Question
Consider the frame and loading shown in the adjoining figure. The temerature acts on the full length of members BC and CD Use the dimensions and applied actions in the tables below. For the combined actions of distributed load and change in temperature, determine the following AT1 1) Horizontal reaction at A 2) Maximum moment in AB For the change in temperature only: 3) Uniform strain component Erin BCD 4) Curvature in member BCD For the combined actions of distributed load and change in temperature 5) Vertical displacement A at point E 6) Relative rotation rel c at point C (Specify the direction of rotation of member CD relative to CB) Enter your answers in the space provided. Dimensions and section lied actions: AT 12 2.2 m 5.2 m 0.3 m 205000 kN EA El 7250| kN 0.00001Explanation / Answer
Let the reaction at A and D be Ax, Ay and Dx , Dy respectively. Let us do separate analysis due to distributed load and temperature load. The results can later be superposed. Only UDL acting Applying force equilibrium. Ax + Dx = w x h w = 10.6 kN/m h = 5.2 m Ax + Dx = 55.12 kN Applying moment equilibrium about A w x h x h/2 = Dx x h Dx = w x h/2 = 27.56 kN Ax = 55.12 -Dx = 27.56 kN Only Temperature load acting As temperatur load is deformation load and member BCD has degree of freedom at B, hence, No axial force and bending moment produced. So horizontal reaction at point A, Ax = 27.56 kN Acting in left direction Since there is hinge at C, the udl will produce moment like a simply supported beam. Maximum moment = w x h^2/8 35.828 kNm Now T1 = -12 degree C T2 = 12.5 degree C = 0.00001 /degree C t = 0.3 m Mean overall rise in temperature = (T1 + T1)*0.5 Tm = 0.25 degree C Uniform strain t = Tm = 2.50E-06 Temperature gradient Tz = (T1 - T1)/t = -81.67 Curvature will be produced by the linear variation of strain with not net strain on the body. equivalent temprature at top and bottom = +/- Tz*t/2 So strain at top and bottom is Top Tt = -12.25 Bottom Tb = 12.25 top = Tt = -1.23E-04 bottom Tb = 1.23E-04 Curvature = (bottom -top)/t = 0.000817 /m So Answers are following 1 27.56 kN 2 35.83 kNm 3 2.50E-06 4 0.000817 /m