Question

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5. The composite beam shown below is made of copper and aluminum. Calculate the maximum normal stress in the copperand in the aluminum due to bending from the applied loads. (15 pts) Ecopper = 117,000,000 kPa Ealuminum = 69,000,000 kPa 30 kN 50 KN 40 KN III Aluminum А E Copper 9 mm 3 mm 24 mm -4 @ 0.675 m = 2.7 m

Answer 1

STEP I: GIVEN 30KN 50KN 40KN A poster tempore 9mm Ę=117x10ʻkra FA= 69x 10 kPa ALUMINIUM COPPER 24mm VE 3 mm 157.5 57.5 27.5 27.5 225 SFDCINKA) 42.5 62.5 57.375 142.1875 138.8125 BMD (IN KNM) 9 5 STEP 1: TO FIND MAXIMUM BENDING STRESS IN ALUMINIUM & COPPER =? STEP TIL: SUPPORT REACTIONS Efx =0 => HA=0 ©<fy =0 =) VA +ve - 30-50 – 40 =0 =) VA+V8 = 120 OEMA=0 - Vp x27–30x0.675 - 50x1:35-40x2.025=0>) Ve=62.5kN SO, VA= 57.5 KN. STEP TV: SFD AB: S fa = VA = 57-5kN S-f& VA = 57.5 KN BC: Sfo=VA -30 = 27.5kN , S. fc = VA-30 = 27.5kN CD: Sf=VA-30-50 = -22.5kN, S.fo=VA --30-50--22:5kN DE: S.fo=VA-30-50-403 - 62.5 KN S.fe=VA- 30-50-403-62-5LN {WE DREW SFD, SO AS TO ENSURE NOT TO MISS ANY POINT WHERE SFD WOULD CHANGE ITS NATURE (FROM +VE TO-VE OR VICE VERSA) AS THIS COULD BE A POINT WHERE MAXIMUM MOMENT OCCURS, FOR GIVEN PROBLEM, WE DO NOT HAVE SUCH A CASE] STEP V:BMD AB: B.MA = VAXD=0 , B.Mg = VAX0.675 = 38.8125kNm BC! B.Mg = VAX0-675= 38.8125kNm B.M = VAXT 35–30x0.675 = 57.375 km CD: B.Mc = VAX1.35-30x0.675 = 57.375 kNmB.Mg=VAX 2.025-30x1.35-50x0.675=424875 DE: BME=V9x2-025–30xl-35-50,0.675=42.1875 kNm B.Mg = VAX27-30x2-025-50x135–40x 0-675= 0 {T TO FIND LOCATION OF MAXIMUM MOMENT BECAUSE AT THIS POINT MAXIMUM BENDING STRESS WILL TAKE PLACE FOR COPPER AND ALUMINIUM 3 kNm IT IS

STEP I: CENTROID OF SECTION TO FIND BENDING STRESSES IN A EASY WAY CONVERT COMPOSTTE SECTION IN TO ONE MATERIAL - SO, LET US CONVERT IT INTO COPPER KEEPING ITS THICKNESS SAME. Ea-7X109-1•6956 9 mm 69X106 3mm 116956x24 = 40.6944 mm 24mm FROM TOP, LET CENTROID LIES AT ye Yo = Ary,+Azyz = (40.6944xg) x (92) +(24x3)x (9+32)5.4857 mm A tA2 (40.69 44x9) + (243) STEP VI! MOMENT OF INERTIA USING PARALLEL AXIS THEOREM, In = 40.6944XgO + (40.694489)x (5. 4857-9.5 = 2828.0346 mmt 52= 34*3*4 (2483)x [5.4857–(9+%)]? = 1964-3107 mit I=I; +Iy = 2828.0346 +1864.3107 = 4692.3453 mmt = My STEP VIL: BENDING STRESS, WE KNOW, I.£ SO, I Я R I 1,= 57.375 x 10°x 5.4857 670756 kPa 4692.3453 02 = 57.375X106x (12-5,4857) 79.6527 kPa 469213453 BUT THESE ARE THE STRESSES IN COPPER. SO, ACTUAL STRESSES IN EACH MATERIAL: PA = 67.0756x 117X106 113.7369 kPa. 69x106 Z ANS Oc= 79.6527x 69X106 79.6527 kPa. -ZANS 69X106