Holooly Plus Logo

Question 17.3: A 4 m long vertical cantilever 100 mm internal diameter stee......

A 4 m long vertical cantilever 100 mm internal diameter steel pipe supports 25 kN weight attached on the top as shown in Fig. 17.16a. D_{o} = 115 ~mm, D_i=100~ mm , I=3.676 \times 10^6~ mm ^4 ~E=200 \times 10^6~ kN / m ^2. Determine the peak deflection; bending stress due to El, Centro ground motion.

fig 17.16
Step-by-Step
The 'Blue Check Mark' means that this solution was answered by an expert.
Learn more on how do we answer questions.

                 k=\frac{3 I}{l^3}~ E

                    =\frac{3 ~\times ~200~\times ~10^6~ \times ~10^3 ~\times ~3.676 ~\times ~10^{-6}}{64}

                    = 34.46 × 10³ N/m

Total weight of the pipe (weight /m = 0.01875 kN/m) = 0.01875 × 4 = 0.075 kN

   Compared with the top weight, the pipe weight is very small and can be neglected.

                 W = 25 kN

                      = 25 000 N

                 m=\frac{25000}{9.81}

                       = 2548 kg

                 \omega_n=\sqrt{\frac{k}{m}}=\sqrt{\frac{34.46~ \times~ 10^3}{2548}}=3.674~ rad / s

                 T_n=\frac{2~ \pi}{\omega_n}=\frac{2~ \pi}{3.674}=1.70 ~s

From Fig. 17.16b

               S_d=127 ~mm

               S_p a / g=0.2

               S_{p a}=V=508 ~mm / s

               f s_0=m S_{p a}=W \times 0.2

                       = 0.2 × 25 = 5 kN

                  \sigma=\frac{M_y}{I}

                     =\frac{5 ~\times ~4~ \times ~10^6~ \times ~115}{2~ \times~ 3.676 ~\times ~10^6}=312~MPa

The stress calculation exceeded the limit, hence the designer decided to increase the size of the pipe as D_0=220~ mm ; D_i=200~mm ; I=3.645 \times 10^7~ mm ^4. Comment on the advantage or disadvantage of using bigger pipes.

                          k = 3EI/l^3

                             =\frac{3 ~\times~ 2 ~\times ~10^8 ~\times ~3.645 \times ~10^{-5}}{64}

                          k = 341.71 kN/mm

                        m = 2548 kg

                      \omega_n=\sqrt{\frac{341.71~ \times ~10^3}{2548}}=11.5 ~rad / s

                      T_n=\frac{2~ \pi}{\omega_n}=0.546

                      S_d=68.58~mm

                 S_{p a / g}=1.11

Horizontal force = 1.11 × 25 = 27.5 kN

                          M = 27.5 × 4 = 111 kN m

                          \sigma=\frac{111~ \times ~10^6 ~\times~ 220}{2~ \times ~3.647~ \times ~10^7}

                             = 334.9 MPa

     The above example points out an important difference between the response of structures due to earthquake excitation and a fixed value of static force. In the static case, the stress decreases by increasing the size of the member. In the case of earthquake excitation the increase in frequency shortens the natural period from 1.7 to 0.54 s which for this spectrum increases the inertia force. Increase or decrease in stress depends on section modulus.

Related Answered Questions