f max=K* 10000kg*300？ /(48 * 2. 1* 10^6* 1 1076)= 1.5 * 10000*27* 10^6(/48* 2. 1* 10^6* 1 1076)=0.3627cm

According to the above calculation of 32 # I-beam, k is 1.5, which exceeds the required deflection by 3.6 times.

Trial calculation with the largest domestic model #63a: change k to1.4;

f max=K* 10000kg*300？ /(48* 2. 1* 10^6*939 16)= 1.4 * 10000*27* 10^6(/48* 2. 1* 10^6*939 16)

= 0.0683cm & lt 1mm meets the requirements.

There seems to be no need to be too strict in strength. Just use Q235.

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