Calculate the slope of elastic line at B and the deflection at point D of the beam shown in Figure 7.29.

Question

Accepted Answer

Let us draw the free-body diagram of the beam as shown in Figure 7.30 to determine the support reactions.

We note from free-body diagram of the beam shown in Figure 7.30(a) that for equilibrium:

[latex] \sum M_{ A }=0 \Rightarrow 20 R_{ B }=(5)(200)+15(120) \Rightarrow R_{ B }=140 kN (\uparrow) [/latex]

and [latex] \sum F_y=0 \Rightarrow R_{ A }=180 kN (\uparrow) [/latex]

In Figure 7.30(b), we show the equivalent loading diagram, which gives

[latex] M_x=(180) x-10 x^2+10\langle x-10 angle^2-120\langle x-15 angle kNm [/latex]

From flexure equation, we get

[latex] (E I) \frac{ d ^2 y}{ d x^2}=-M_x=-180 x+10 x^2-10\langle x-10 angle^2+120\langle x-15 angle [/latex]

Now integrating successively, we obtain:

[latex] (E I)\left\lgroup\frac{ d y}{ d x} ight group =-90 x^2+\frac{10}{3} x^3-\frac{10}{3}\langle x-10 angle^3+60\langle x-15 angle^2+C_1 [/latex] (1)

and [latex] (E I) y=-30 x^3+\frac{10}{12} x^4-\frac{10}{12}\langle x-10 angle^4+20\langle x-15 angle^3+C_1 x+C_2 [/latex] (2)

Noting the boundary conditions of y = 0 at x = 0 m and at x = 20 m, we get from Eq. (2) that:

[latex] C_2=0 \quad ext { and } \quad C_1=5625 kN m ^2 [/latex]

Therefore Eqs. (1) and (2) become

[latex] (E I) \frac{ d y}{ d x}=-90 x^2+\frac{10}{3} x^3-\frac{10}{3}\langle x-10 angle^3+60\langle x-15 angle^2+5625 [/latex] (3)

and [latex] (E I) y=-30 x^3+\frac{10}{12} x^4-\frac{10}{12}\langle x-10 angle^4+20\langle x-15 angle^3+5625 x [/latex] (4)

Putting x = 20 m in Eq. (3), we get

[latex] \left.\frac{ d y}{ d x} ight|_{x=20 m }=\left.\frac{ d y}{ d x} ight|_{ B }= an heta_{ B } \approx heta_{ B }=-\frac{5541.67}{E I}=-\frac{5541.67}{(200)\left(10^6 ight)(2.5)\left(10^{-12} ight)\left(10^9 ight)} [/latex]

Therefore,

[latex] heta_{ B }=-0.0111 rad [/latex]⦨

Again, putting x = 15 m in Eq. (4), we obtain

[latex] ext { (EI) }\left.y ight|_{x=15 m }=\left.24791.67 \Rightarrow y ight|_{x=15 m }=\delta_{ D }=\frac{24791.67}{(200)\left(10^6 ight)(2.5)\left(10^9 ight)\left(10^{-12} ight)} m [/latex]

Therefore,

[latex] \delta_{ D }=0.04958 m =49.58 mm [/latex]

Question 7.16: Calculate the slope of elastic line at B and the deflection ...

Related Answered Questions

For the previous problem, calculate the deflection at point C on the beam. ...

Verified Answer:

For the beam shown in Figure 7.31, find the support reactions. Assume EI to be constant. ...

Verified Answer:

Calculate the deflections at points A and C of the beam shown in Figure 7.27. Assume EI to be constant. ...

Verified Answer:

For the beam with EI constant shown in Figure 7.23, obtain the equation for the elastic line and determine the slope of the elastic line at the right end of the beam. ...

Verified Answer:

A simple beam AB is subjected to couples Mo and 2Mo acting as shown in Figure 7.25. Calculate the angles of rotation θA and θB and maximum deflection, δmax. Assume EI to be constant. ...

Verified Answer:

Determine the angle of rotation θB and deflection δB at the free end of a cantilever beam AB with uniform load intensity wo as shown in Figure 7.21. Consider EI to be constant. ...

Verified Answer:

For the beam shown in Figure 7.19, compute (a) the slope at A and (b) deflection at C. Assume EI to be constant. ...

Verified Answer:

A simply supported beam AB carries a concentrated load P at the midpoint C. The left half of the beam has area moment of inertia I while the right half has area moment of inertia 2I. Find out slopes at A and B and deflection at C [(refer Figure 7.15(a)]. ...

Verified Answer:

For a simply supported beam as shown in Figure 7.14 find out the slope at point Q if Young’s modulus is 200 GPa and area moment of inertia is 17 × 10^6 mm^-4. ...

Verified Answer:

Using moment-area method find out the slope at point C in the Figure 7.13(a). ...

Verified Answer: