Question 1 25 points a wood beam ab supporting two...

A wood beam AB supporting two concentrated loads P (see Fig:) has a rectangular cross-section of width b = 130 mm and height h = 150 mm (see Fig: b). The distance from each end of the beam to the nearest load is a = 1.0 m. Determine the maximum permissible value Pmax of the loads if the allowable stress in bending is Oallow = 12 MPa and the allowable stress in horizontal shear is Tallow = 1.2 MPa. (Disregard the weight of the beam itself)

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00:01 Hello everyone in this question it is given that the allowable stress in bending is given by sigma allowable equals to 12 megapascal and the value of allowable stress stress which you can retain as tau allowable equals to 1 .2 megabaskar so here we need to find out the value of maximum flow so the maximum safe force occurred at the support and the bending moment occurs throughout the region between the load so therefore from here you can as v suffix max equals to p so from n here you can retain as for bending moment m suffix max equals to p multiplied by a so the section modulus s and constitutional area you can t as s equals to i devolved by c and here the value of i is v hq divided by 12 and the value of c is h divided by 2 so further solving this from here we will got the value of section modulus s equals to b a c squared divided by 6 and the value of area here we can attain as area a equals to width b multiplied by high h so the maximum normal stress can be obtained by using taxo formula you can maintain as the sextro formula equals to sigma max equals to bending moment max divided by xxon modulus s so now here we will put the values so therefore from here we will get sigma max equals to 6a divided by b ac squared so the series stress in the beam are obtained by using the formula again you can return the formula for series stress which is tau max equals to c multiplied by g max divided by two cross -section area a so now here we will put the values so therefore from here we will get tau max equals to cp divided by 2 b multiplied by h so therefore the maximum permissible values of the load p in bending and shared respectively you can attain as the formula t suffix b 4 bending which is equal to sigma allowable multiplied by b s squared divided by 6a so now here we will put the values so after putting the values we will go the mix load for bending c suffix b equals to 12 multiplied by 130 mm multiplied by 150 m whole squared divided by 6 multiplied by so therefore after putting the values we will got the value of p suffix b for means load for bending equals to 5 .85 kilo neuter and now for zero load we can retain as p suffix as means this is p suffix c equals to 2 multiplied by tau allowable multiplied by b bx divided by 3 so now here we will put the values so after putting the value we will go the value of share load equals to 15 .6 kilo newton but here in this case bending mistress gober under design and the maximum permissible load you can retain as here p suffix max equals to 5 .85 kilo newton so therefore from here you can retain as our final answered maximum permissible load in suffix max equals to 5 .85 kilo neither...

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