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Lec6 MECH ENG STRucture
- 1. 2.001 - MECHANICS AND MATERIALS I Lecture #6 9/27/2006 Prof. Carol Livermore Recall: Sign Convention Positive internal forces and moments shown. Distributed Loads 1
- 2. q(x) = q0 L L FTotal Loading = q(x)dx = q0 x = q0 L 0 0 L L q0 x2 q0 L2 MADistributed Load = q(x)xdx = = 0 2 0 2 Lump: (Equivalent Forces) FBD 2
- 3. Fx = 0 R Ax = 0 Fy = 0 RAy + RBy q0 L = 0 MA = 0 L q 0 L + RBy L = 0 2 q0 L R By = 2 Substitute: q0 L R Ay + q0 L = 0 2 qo L R Ay = 2 Cut beam. 3
- 4. Lump: Fx = 0 N =0 Fy = 0 q0 L L q 0 x Vy = 0 Vy q 0 x 2 2 M = 0 q0 L x x + q0 x +Mz = 0 2 2 L x2 Mz = q0 x 2 2 Plot
- 5. Sanity Check Pinned and Pinned on rollers at end Cannot support moment M at ends Can support x and y loads OK Fy = 0 Vy (Vy + 隆Vy ) + q隆x = 0 隆Vy = q隆x qdVy Limit as 隆x 0 dx M0 = 0 隆x 隆x Mz + Mz + 隆Mz Vy (Vy + 隆Vy ) =0 2 2 隆x 隆x 隆Mz 2Vy 隆Vy 2 2 隆x 隆Mz = Vy 隆x 隆Vy 2 隆Mz 隆Vy Vy = + 隆x 2 Take limit 隆x 0 dMz Vy = dx dVy q= dx But what if load is not uniformly distributed?
- 6. EXAMPLE: A more interesting structure Q: Find all internal forces and moments in all members. Plot. Find reactions at supports. FBD Fx = 0 RAx + RDx = 0 Fy = 0 RAy + RDy = 0 MA = 0 M RDx L + RDy L = 0 FBD
- 7. This is a 2-force member. L R Ay 1 = tan 慮 = 2 = R Ax L 2 R Ax RAx = 2RAy =2 R Ay R Ax = 1 R Dx R Ay = 1 R Dy R Ax R Ay = R Dx R Dy R Ax R Dx = R Ay R Dy So: R Ax = 2 RDx = 2RDy R Dy So: M R Dy = L M R Ay = L 2M R Dx = L 2M R Ax = L Fx = 0 R Ax R C x = 0 2M RCx = L Fy = 0 R Ay R C y = 0 M R Cy = L 7
- 8. Take cut FBD Fx = 0 2M N =0 L 2M N= L Fy = 0 M Vy = 0 L M Vy = L M = 0 M x + Mz = 0 L M Mz = x L FBD of Cut 2 8
- 9. Fx = 0 M +N =0 L M N= L Fy = 0 2M Vy = 0 L 2M Vy = L M = 0 M 2M Mz + x x=0 L L 2x Mz = M 1 L FBD of Cut 3 Fx = 0 M +N =0 L M N= L Fy = 0 2M Vy = 0 L 9
- 10. 2M Vy = L M = 0 2M Mz + x=0 L 2M Mz = x L FBD of Cut 4 Fx = 0 M +N =0 L M N= L Fy = 0 2M Vy = 0 L 2M Vy = L M = 0 2M M x + Mz = 0 L 2x Mz = M 1 L 10