A single 6 -mm-diameter steel pin B is used to connect the...

A single 6 -mm-diameter steel pin $B$ is used to connect the steel strip $D E$ to two aluminum strips, each of 20 -mm width and $5-\mathrm{mm}$ thickness. The modulus of elasticity is 200 GPa for the steel and 70 GPa for the aluminum. Knowing that for the pin at $B$ the allowable shearing stress is $\tau_{\mathrm{all}}=85 \mathrm{MPa}$, determine, for the loading shown, the maximum strain energy that can be acquired by the assembled strips.

Transcript

00:01 We are given a 6mm diameter diameter so we are given here 6mm diameter diameter of the steel pin so steel pin we are given you have a diameter steel pin b have a diameter of 6mm now the strip de we are given here strip d to the aluminum strip and each of a 20 meter weight so b which is equal we are given a 20 meter to any millimeter width and the thickness we are given a t which is equal to 5 million millimeter thickness and the modulus of elasticity here we can write modules of elasticity that is we can say the row which is equal to that is 20 giga pascal not 20 it is 200 giga pascal for the steel and the is for the scale and this we are given here that 70 giga pascal for the aluminum and now here we can try the here the allowable seeding stress that is t of all taw all which is equal we can make 85 megapascal now determine the loading son and maximum strain energy you that we want to find here so let's solve this question and finding the answer the first we can write here the diameter of the still 6 meter which we are given here here with the 20 and the thickness we are given or with the thickness also denoted by d and diameter 6 minute that's 200 we are given elasticity 70 and that's we are given all over steering stress that is 85 megapascal so let's solve this question first we can write here that eliminate our area a which is equal to b multiplied by d here and here we can write it substituting values here so we can write here answer is hundred millimeter square now here we know that area of pin here area of pin which is equal we can write pi divided by 4 into the bracket we can write d pina square now we substitute value into this formula we are given here 6mm not 6 is 5 millimeter we are given here we are substituting value into this formula so we can write here the answer is 28 .27 millimeter square now here we can write defining the value of length b so length b which is equal we can write 1 .25 invite 1 .25 minus we can write 0 .5 which is equal answer is 0 .5 which is equal answer is 0 .5 7 5 meter and define the force p so here we can write the force p which is equal to 2 multiply by a pin multiple we can write the tau all multiple 12 all pin and now substituting the value into this formula so here we can write the value here is the p multiple by 2 where we can write 28 point 274 malive we can write here the value is 8 value here is the 85 and the we just simplify and we can write here answer is so here you can write 2 multidavre by 28 .274 by 85 here we can write 85 which is simply write and we can write here answer is 4 ,800 and 880s newton write pab which is equal we can write bdb and now here we can write b b and now here we can write the pab which…

Question

Please give Ace some feedback